Question

In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the A.P.

Answer 1

use formula
Sn=n/2 {2a+(n-1) d}
where a is first term of AP and d is common difference
S10=10/2 {2a+(10-1) d}
210=5 (2a+9d)
2a+9d=42 -------------(1)
now again according to question
last 15 terms sum =2565
S50-S35=2565
50/2 {2a+49d}-35/2 (2a+34d)=2565
25 (2a+49d)-35 (a+17d)=2565
50a+1225d-35a-595d=2565
15a+630d=2565
a+62d=171 -------------(2)
solve equation (1) and (2) and find a and d also find AP

Answer 2

$\underline{ \underline{\bf{Answer}}} : - \\ \implies \: 3, \: 7 \:, 11 \: ,15, \: ..........,199 \\ \\ \underline{\underline{ \bf{Step - by - step \: explanation \: }}} : -$

According to the question:-

$\bf{sum \: of \: first \: 10 \: terms \:( s_{10}) = 210} \\ 210 = \frac{10}{2} \bigg (2a + (101)d \bigg) \: \\ \\ 2a + 9d = 42 \: .........(1)\\ \\ \bf{sum \: of \: last \: 15 \: terms \: ( s_{15})= 2565} \\ \\ s_{50} -s_{35} = 2565 \\ \\ 2565 = \frac{50}{2} \bigg(2a + (50 - 1)d \bigg) - \frac{35}{2} \bigg(2a + (35 - 1)d \bigg) \\ \\ 2565 = 25(2a + 49d) - 35(a + 17d) \\ \\ 2565 = 50a + 1225d - 35a - 595d \\ \\ after \: solving \: this \: \\ \\ a + 42d = 171 \: ...........(2) \\ \\ from \: eq(1) \: and \: (2) \\ \\eq (1) \times 42 - \: eq (2) \times 9 \\ \\ we \: get \: \\ \\ a = 3 \: d = 4$

Hence required AP is →

3,7,11,15,....,199