Question

In an AP of 50 terms the sum of first 10 term is 210 and the sum of its last 15 terms is 2565 find the AP

Answer 1

sum of first 10 terms is 210
therefore
S10=10/2(2a+(10-1)d)=210
5(2a+9d)=210
2a+9d=42.....(1)

NOW THE SUM OF ITS LAST 15 TERMS IS 2565
THEREFORE
15 th term from the last = ( 50 – 15 + 1 ) th = 36 th term from the beginning

⇒ a36 = a + 35d
S15=15/2(2a+(15-1)d)=2565
15/2(2a+14d)=2565
2a+14d=342
2(a+35d)+14d=342
a+42d=171.....(2)
from (1) and (2) we get a=3 and d=4
now 50th term is
a+49d
3+49(4)
199=T50
therefore your ap is 3,7,11,15.....199

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