Question

40. Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find
41 Find the sum of first 51 terms of an AP whose second and third term
obtained
25th term.
(CBSE 2​

Answer 1

S14=1050

⇒n/2[2a+(n-1)d]=1050

⇒14/2[20+13d]=1050

⇒7[20+13d]=1050

⇒20+13d=1050/7

⇒20+13d=150

⇒13d=150-20

⇒13d=130

⇒d=10

20th term=a+19d

                =10+19×10

                =10+190

                =200