Advertisements
Advertisements
Question
Find the sum of first 16 terms of the A.P. whose nth term is given by an = 5n – 3.
Solution
Given, nth term of A.P. is an = 5n – 3
∴ a1 = 5(1) – 3 = 2
And a2 = 5(2) – 3 = 7
∴ Common difference, d = 7 – 2 = 5
∴ Sum of n terms of A.P., Sn = `n/2[2a + (n - 1)d]`
∴ S16 = `16/2 [2(2) + (16 - 1)5]`
= 8(4 + 75)
= 8 × 79
= 632
RELATED QUESTIONS
If the term of m terms of an A.P. is the same as the sum of its n terms, show that the sum of its (m + n) terms is zero
Show that a1, a2,..., an... form an AP where an is defined as below:
an = 9 − 5n
Also, find the sum of the first 15 terms.
Is -150 a term of the AP 11, 8, 5, 2, ……?
How many two-digits numbers are divisible by 3?
The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its
Find S10 if a = 6 and d = 3
In an AP if a = 1, an = 20 and Sn = 399, then n is ______.
In an A.P., the sum of first n terms is `n/2 (3n + 5)`. Find the 25th term of the A.P.
If the first term of an A.P. is p, second term is q and last term is r, then show that sum of all terms is `(q + r - 2p) xx ((p + r))/(2(q - p))`.