Question

An arithmetic progression has first term a and common difference d. It is given that the sum of the first 200 terms is 4 times the sum of the first 100 terms.

i) find d in terms of a
ii)find the 100 th term in terms of a

Ans:i) d=2a ii) T100 = 199a ​

Answer 1

Given:

AP with first term a and common difference d

Sum of first 200 terms of AP = 4 (Sum of first 100 terms of AP)

To find:

(i) The common difference (d) in terms of a

(ii) The 100th term in terms of a

Solution:

Sum of first n terms of AP=    $\frac{n}{2} [2a+(n-1)d]$

According to the question,

⇒$\frac{200}{2} (2a+199d) = 4 [\frac{100}{2} (2a+99d)]$

or 2a + 199d = 4a + 198d

or d=2a

⇒nth term = a+(n-1)d

so 100th term = a+99d

= a+99x2a

=199a

Hence, d = 2a and the 100th term is 199a.