Класс 10 RD Sharma Solutions – Глава 9 Арифметические прогрессии – Упражнение 9.6 | Набор 2
Вопрос 25. В АП первый член равен 22, n-й член равен –11, а сумма первых n членов равна 66. Найдите n и d, общую разность.
Решение:
Given A.P. has first term(a) = 22, nth term(an) = –11 and sum(Sn) = 123.
Now by using the formula of sum of n terms of an A.P.
Sn = n[a + an] / 2
=> 66 = n[22 + (−11)]/2
=> 66 = n[22 − 11]/2
=> 11n = 132
=> n = 12
By using the formula of nth term of an A.P.
an = a + (n – 1)d
=> −11 = 22 + (12 – 1)d
=> 11d = –33
=> d = –3
Hence, the number of terms of given A.P. is 12 and common difference is –3.
Вопрос 26. Первый и последний члены АП равны 7 и 49 соответственно. Если сумма всех его членов равна 420, найдите общую разность.
Решение:
Given A.P. has first term(a) = 7, last term(an) = 49 and sum(Sn) = 420.
Now by using the formula of sum of n terms of an A.P.
Sn = n[a + an] / 2
So,
=> 420 = n[7 + 49]/2
=> 28n = 420
=> n = 15
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
=> 49 = 7 + (15 – 1)d
=> 14d = 42
=> d = 3
Hence, the common difference of the given A.P. is 3.
Вопрос 27. Первый и последний члены АП равны 5 и 45 соответственно. Если сумма всех его членов равна 400, найдите его общую разность.
Решение:
Given A.P. has first term(a) = 5, last term(an) = 45 and sum(Sn) = 400.
Now by using the formula of sum of n terms of an A.P.
Sn = n[a + an] / 2
=> 400 = n[5 + 45]/2
=> 25n = 400
=> n = 16
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
=> 45 = 5 + (16 – 1)d
=> 15d = 40
=> d = 8/3
Hence, the common difference of the given A.P. is 8/3.
Вопрос 28. Сумма первых 9 членов АП равна 162. Отношение его 6-го члена к 13-му члену равно 1:2. Найдите первый и 15-й члены АП.
Решение:
We know sum of n terms of an A.P. is given by Sn = n[2a + (n − 1)d] / 2.
Therefore, Sum of first 9 terms of given A.P. = S9 = 9[2a + (9 − 1)d] / 2 = 162
=> 162 = 9(2a + (9 − 1)d) / 2
=> 2a + 8d = 36
=> a + 4d = 18 …..(1)
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
Given a6 : a13 = 1 : 2,
=> a13 = 2a6
=> a+12d = 2(a + 5d)
=> a+12d = 2a + 10d
=> a = 2d …..(2)
On putting (2) in (1), we get,
=> 2d + 4d = 18
=> 6d = 18
=> d = 3
On putting d = 3 in (2), we get,
a = 2(3) = 6, which is the first term.
Now 15th term, a15 = a + 14d = 6 + 14 × 3 = 6 + 42 = 48
Hence, the first and 15th term of the A.P. are 6 and 48 respectively.
Вопрос 29. Если 10-й член АП равен 21, а сумма его первых 10 членов равна 120, найдите его n-й член.
Решение:
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
10th term of the given A.P., a10 = 21
=> a + 9d = 21 …..(1)
By using the formula of the sum of n terms of an A.P.
Sn = n[2a + (n − 1)d] / 2.
So,
S10 = 10[2a + (10 − 1)d] / 2
=> 120 = 5(2a + 9d)
=> 2a + 9d = 24 …. (2)
On subtracting (1) from (2), we get
=> 2a + 9d – a – 9d = 24 – 21
=> a = 3
On putting a = 3 in eq(1), we get,
=> 3 + 9d = 21
=> 9d = 18
=> d = 2
So, an = 3 + (n – 1)2
= 3 + 2n – 2
= 2n + 1
Hence, the nth term of the given A.P is 2n + 1.
Вопрос 30. Сумма первых 7 членов АП равна 63, а сумма следующих 7 его членов равна 161. Найдите 28-й член этого АП.
Решение:
Sum of first 7 terms of an A.P., S7 = 63.
And sum of next 7 terms is 161.
So, the sum of first 14 terms, S14 = Sum of first 7 terms + Sum of next 7 terms
S14 = 63 + 161 = 224
By using the formula of the sum of n terms of an A.P.
Sn = n[2a + (n − 1)d] / 2.
So, S7 = 7(2a + (7 − 1)d) / 2
=> 7(2a + 6d) / 2 = 63
=> 2a + 6d = 18 . . . . (1)
Also, S14 = 14(2a + (14 − 1)d) / 2
=> 14(2a+13d)/2 = 224
=> 2a+13d = 32 . . . . (2)
Now, subtracting eq(1) from eq(2), we get
=> 13d – 6d = 32 – 18
=> 7d = 14
=> d = 2
On putting d = 2 in eq(1), we get,
=> 2a + 6(2) = 18
=> 2a = 18 – 12
=> a = 3
Thus, a28 = a + (28 – 1)d = 3 + 27 (2) = 3 + 54 = 57
Hence, the 28th term is 57.
Вопрос 31. Сумма первых семи членов АП равна 182. Если его 4-й и 17-й члены относятся как 1:5, найти АП
Решение:
By using the formula of the sum of n terms of an A.P.
Sn = n[2a + (n − 1)d] / 2.
So, S7 = 7(2a + (7 − 1)d) / 2
=> 7(2a + 6d) = 364
=> 14a + 42d = 364
=> a + 3d = 26 .… (1)
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
We are given, a4 : a17 = 1:5
=> a17 = 5a4
=> a+16d = 5[a + 3d]
=> a + 16d = 5a + 15d
=> 4a = d …. (2)
Using eq(2) in eq(1), we get,
=> a + 3(4a) = 26
=> 13a = 26
=> a = 2
On putting a = 2 in eq(2), we get,
=> d = 8
As the first term of the given A.P. is 2 and
the common difference is 8, So, the A.P. is 2, 10, 18, 26, …..
Вопрос 32. N-й член АП равен (−4n + 15). Найдите сумму первых 20 членов этой АП.
Решение:
We are given the nth term of the A.P., an = −4n + 15.
On putting n = 1 to find the first term of our A.P., a = −4(1) + 15 = 11
On putting n = 20 to find the 20th term of the A.P., a20 = −4(20) + 15 = −65
By using the formula of the sum of first n terms of an A.P.
Sn = n[a + an] / 2.
So, S20 = 20[11 + (−65)] / 2
= 10(−54) = −540
Hence, the sum of first 20 terms of this A.P. is −540.
Вопрос 33. В АП сумма первых десяти членов равна −150, а сумма следующих 10 членов равна −550. Найдите точку доступа
Решение:
Sum of first ten terms, S10 = −150.
Also given, Sum of its next 10 terms = −550
Sum of first 20 terms, S20 = Sum of first 10 terms + Sum of next 10 terms
=> S20 = −150 + (−550) = −700
By using the formula of the sum of n terms of an A.P.
Sn = n[2a + (n − 1)d] / 2.
So, S10 = 10(2a + (10 − 1)d) / 2
=> −150 = 5(2a + 9d)
=> 2a + 9d = –30 ….. (1)
And also, S20 = 20(2a + (20 − 1)d) / 2
=> −700 = 10(2a + 19d)
=> 2a + 19d = −70 ….. (2)
Now, subtracting eq(1) from (2), we get
=> 19d – 9d = –70 – (–30)
=> 10d = –40
=> d = –4
On putting d = –4 in (1), we get,
=> 2a + 9(–4) = –30
=> 2a = 6
=> a = 3
As we have a = 3 and d = –4, hence, the A.P is 3, –1, –5, –9,…..
Вопрос 34. Сумма первых 14 членов АП равна 1505, а его первый член равен 10. Найдите его 25-й член.
Решение:
Given A.P. has first term(a) = 10 and sum of the first 14 terms, S14 = 1505.
By using the formula of the sum of n terms of an A.P.
Sn = n[2a + (n − 1)d] / 2.
=> S14 = 14(2(10) + (14 − 1)d) / 2 = 1505
=> 7(20 + 13d) = 1505
=> 20 + 13d = 215
=> 13d = 195
=> d =15
We know the 25th term is, a25 = 10 + (25 − 1)15
= 10 + 24(15)
= 370
Hence, the 25th term of the A.P is 370.
Вопрос 35. В АП первый член равен 2, последний член равен 29, а сумма членов равна 155. Найдите общую разность АП
Решение:
Given A.P. has first term(a) = 2, last term(an) = 29 and sum(Sn) = 155.
By using the formula of the sum of first n terms of an A.P.
Sn = n[a + an] / 2.
=> 155 = n(2 + 29) / 2
=> 31n = 310
=> n = 10
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
=> 29 = 2 + (10 – 1)d
=> 9d = 27
=> d = 3
Hence, the common difference of the A.P. is 3.
Вопрос 36. Первый и последний члены АП равны 17 и 350 соответственно. Если общая разность равна 9, сколько членов и какова их сумма?
Решение:
Given A.P. has first term(a) = 17, common difference(d) = 9 and last term(an) = 350.
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
=> 350 = 17 + (n – 1) 9
=> 350 = 17 + 9n – 9
=> 9n = 342
=> n = 38
By using the formula of the sum of first n terms of an A.P.
Sn = n[a + an] / 2.
S38 = 38(17 + 350)/2
= 19(367) = 6973
Hence, the number of terms of the given A.P is 38 and sum is 6973.
Вопрос 37. Найдите количество членов АП –12, –9, –6, . . . , 21. Если к каждому члену этой АП прибавить 1, то найти сумму всех членов полученной таким образом АП.
Решение:
If 1 is added to each term of the A.P. then the new A.P. is –11, –8, –5, . . . , 22.
First term, a = –11 and common difference, d = – 8 – (–11) = 3
And, we know that nth term = an = a + (n – 1)d
=> 22 = –11 + (n – 1)3
=> 3n = 36
=> n = 12
By using the formula of the sum of first n terms of an A.P.
Sn = n[a + an] / 2.
=> S12 = 12[–11 + 22]/2
= 6[11]
= 66
Hence, the sum after adding 1 to each of the terms in the A.P is 66.
Вопрос 38. Сумма первых n членов АП равна 3n 2 + 6n. Найдите n-й член этой АП
Решение:
Given Sn = 3n2 + 6n
On putting n = 1, we get the first term(a), S1 = a = 3(1)2 + 6(1) = 9
On putting n = 2 gives S2 = a + a + d = 3(2)2 + 6(2) = 24
=> d = 24 – 2a
=> d = 24 – 18 = 6
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
= 9 + (n – 1)6
= 9 + 6n – 6
= 6n + 3
Hence, the nth term of the given A.P. is 6n + 3.
Вопрос 39. Сумма первых n членов АП равна 5n – n 2 . Найдите n-й член этой АП
Решение:
Given Sn = 5n – n2,
On putting n = 1, we get the first term(a), S1 = a = 5(1) – (1)2 = 4
On putting n = 2 gives S2 = a + a + d = 5(2) – (2)2 = 6
=> d = 6 – 2a
=> d = 6 – 8 = –2
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
= 4 + (n – 1)(–2)
= 4 – 2n + 2
= 6 – 2n
Hence, the nth term of the given A.P. is 6 – 2n.
Вопрос 40. Сумма первых n членов АП равна 4n 2 + 2n. найти n-й член этой АП
Решение:
Given Sn = 4n2 + 2n,
On putting n = 1, we get the first term(a), S1 = a = 4(1)2 + 2(1) = 6
On putting n = 2 gives S2 = a + a + d = 4(2)2 + 2(2) = 20
=> d = 20 – 2a
=> d = 20–12 = 8
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
= 6 + (n – 1)8
= 6 + 8n – 8
= 8n – 2
Hence, the nth term of the given A.P. is 8n – 2.
Вопрос 41. Сумма первых n членов АП равна 3n 2 + 4n. найти 25-й член этой АП
Решение:
Given Sn = 3n2 + 4n,
On putting n = 1, we get the first term(a), S1 = a = 3(1)2 + 4(1) = 7
On putting n = 2 gives S2 = a + a + d = 3(2)2 + 4(2) = 20
=> d = 20 – 2a
=> d = 20–14 = 6
The 25th term of the A.P., a25 = a + (25 – 1)d
= 7 + (24)8
= 7 + 144
= 151
Hence, the 25th term of the A.P. is 151.
Вопрос 42. Сумма первых n членов АП равна 5n 2 + 3n. Если его m -й член равен 168, найдите значение m. Кроме того, найдите 20-й член этой АП
Решение:
Given Sn = 5n2 + 3n,
On putting n = 1, we get the first term(a), S1 = a = 5(1)2 + 3(1) = 8
On putting n = 2 gives S2 = a + a + d = 5(2)2 + 3(2) = 26
=> d = 26 – 2a
=> d = 26 – 16 = 10
The mth term of the A.P., am = a + (m – 1)d = 168
=> 8 + (m – 1)10 = 168
=> (m – 1)10 = 160
=> m – 1 = 16
=> m = 17
20th term of the A.P., a20 = a + 19d
= 8 + 19(10)
= 8 + 190
= 198
Hence, the value of m is 17 and 20th term of the A.P. is 198.
Вопрос 43. Сумма первых q членов АП равна 63q – 3q 2 . Если его p -й член равен –60, найдите значение p. Кроме того, найдите 11-й член этой АП.
Решение:
Given Sn = 63q – 3q2,
On putting n = 1, we get the first term(a), S1 = a = 63(1) – 3(1)2 = 60.
On putting n = 2 gives S2 = a + a + d = 63(2) – 3(2)2 = 114
=> d = 114 – 2a
=> d = 114 – 120 = –6
The pth term of the A.P., ap = a + (p – 1)d = –60
=> 60 + (p – 1)(–6) = –60
=> (p – 1)(–6) = –120
=> p – 1 = 20
=> p = 21
11th term of the A.P., a11 = a+10d
= 60 + 10(–6)
= 0
Hence, the value of p is 21 and 11th term of the A.P. is 0.
Вопрос 44. Сумма первых m членов АП равна 4m 2 – m. Если его n-й член равен 107, найдите значение n. Кроме того, найдите 21-й член этой АП.
Решение:
Given Sm = 4m2 – m,
On putting m = 1, we get the first term(a), S1 = a = 4(1)2 – 1 = 3.
On putting m = 2 gives S2 = a + a + d = 4(2)2 – 2 = 14
=> d = 14 – 2a
=> d = 14 – 6 = 8
The nth term of the A.P., an = a + (n – 1)d = 107
=> 3 + (n – 1)8 = 107
=> 8(n – 1) = 104
=> n – 1 = 13
=> n = 14
21st term of the A.P., a21 = a + 20d
= 3 + 20(8)
= 163
Hence, the value of n is 14 and 21st term of the A.P. is 163.
Вопрос 45. Если сумма первых n членов АП равна 4n – n 2 , то чему равен первый член? Чему равна сумма первых двух членов? Какой второй срок? Аналогичным образом найдите 3-й, 10-й и n-й члены.
Решение:
Given Sn = 4n – n2,
On putting n = 1, we get the first term(a), S1 = a = 4(1) – 12 = 3.
On putting n = 2 gives S2 = a + a + d = 4(2) – 22 = 4
=> d = 4 – 2a
=> d = 4 – 6 = –2
Second term(a2) = a + d = 3 + (–2) = 1
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
= 3 + (n – 1)(–2)
= 5 – 2n
Hence, third term(a3) = 5 – 2(3) = –1
And tenth term(a10) = 5 – 2(10) = –15
Hence, the first term is 3, sum of first two terms is 4, second term is 1
and 3rd, 10th and nth terms are –1, –15 and 5 – 2n respectively.
Вопрос 46. Если сумма первых n членов АП равна 
затем найдите его n-й член. Следовательно, напишите его 20-й член.
Решение:
Given Sn = (3n2 + 7n) / 2,
On putting n = 1, we get the first term(a), S1 = a = [3(1)2 + 7(1)]/2 = 10/2 = 5
On putting n = 2 gives S2 = a + a + d = [3(2)2 + 7(2)]/2 = 26/2 = 13
=> d = 13 – 2a
=> d = 13 – 10 = 3
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
= 5 + (n – 1)3
= 5 + 3n – 3
= 3n + 2
Now we find the 20th term:
a20 = 3(20) + 2
= 60 + 2
= 62
Hence, the nth term of the given A.P. is 3n + 2 and the 20th term is 62
Вопрос 47. В АП сумма первых n слагаемых равна 
. Найдите его 25-й член.
Решение:
Given Sn = 3n2/2 + 13n/2,
On putting n = 1, we get the first term(a), S1 = a = 3(1)2/2 + 13(1)/2 = (3 + 13) / 2 = 8
On putting n = 2 gives S2 = a + a + d = 3(2)2/2 + 13(2)/2 = 19
=> d = 19 – 2a
=> d = 19 – 16 = 3
The 25th term of the A.P., a25 = a + (25 – 1)d
= 8 + (24)3
= 6 + 72
= 80
Hence, the 25th term of the given A.P. is 80.
Вопрос 48. Найдите сумму всех натуральных чисел от 1 до 100, которые делятся на 3.
Решение:
Natural numbers between 1 and 100 which are divisible by 3 are 3, 6, 9, 12, . . . 99.
These numbers form an A.P. with first term(a) = 3,
Common difference(d) = 6 – 3 = 3 and nth term(an) = 99.
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
=> 99 = 3 + (n – 1)3
=> 3(n – 1) = 96
=> n – 1 = 32
=> n = 33
By using the formula of the sum of first n terms of an A.P.
Sn = n[a + an] / 2.
So,
S33 = 33[3 + 99]/2
= 33[51]
= 1683
Hence, the sum of all natural numbers between 1 and 100 which are divisible by 3 is 1683.