Advertisements
Advertisements
Question
Calculate the sum of 35 terms in an AP, whose fourth term is 16 and ninth term is 31.
Advertisements
Solution
Let a and d be the first term and the common difference, respectively.
Now, a4 = 16 ......[Given]
⇒ a + 3d = 16 .....(i)
Also, a9 = 31 ......[Given]
⇒ a + 8d = 31 .....(ii)
Subtracting the equation (i) from equation (ii), we get
a + 8d = 31
a + 3d = 16
– – –
5d = 15
⇒ d = 3
Substituting the value of d in equation (i), w e get
a + 3 (3) = 16
⇒ a + 9 = 16
⇒ a = 16 – 9 = 7
Now, `"S"_n = n/2 [2a + (n - 1)d]`
∴ `"S"_35 = 35/2 [2(7) + (35 - 1)3]`
= `35/2 [14 + (34)3]`
= `35/2 (14 + 102)`
= `35/2 (116)`
= 35 × 58
= 2030
Hence, the sum of 35 terms of the A.P. is 2030.
APPEARS IN
RELATED QUESTIONS
If the sum of the first n terms of an A.P. is `1/2`(3n2 +7n), then find its nth term. Hence write its 20th term.
Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120
The sum of the first p, q, r terms of an A.P. are a, b, c respectively. Show that `\frac { a }{ p } (q – r) + \frac { b }{ q } (r – p) + \frac { c }{ r } (p – q) = 0`
In an AP given a = 3, n = 8, Sn = 192, find d.
Which term of the AP 21, 18, 15, …… is -81?
Find the middle term of the AP 10, 7, 4, ……., (-62).
The 9th term of an AP is -32 and the sum of its 11th and 13th terms is -94. Find the common difference of the AP.
What is the 5th term form the end of the AP 2, 7, 12, …., 47?
Write an A.P. whose first term is a and common difference is d in the following.
a = –1.25, d = 3
In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.
(Assume that three consecutive terms in A.P. are a – d, a, a + d).
If m times the mth term of an A.P. is eqaul to n times nth term then show that the (m + n)th term of the A.P. is zero.
The sum of first n terms of an A.P. is 5n − n2. Find the nth term of this A.P.
If the sum of n terms of an A.P. is 2n2 + 5n, then its nth term is
If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 terms is
The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is
If a = 6 and d = 10, then find S10
If 7 times the seventh term of the AP is equal to 5 times the fifth term, then find the value of its 12th term.
If the first term of an A.P. is 5, the last term is 15 and the sum of first n terms is 30, then find the value of n.
The sum of 40 terms of the A.P. 7 + 10 + 13 + 16 + .......... is ______.
