Question

find the sum of 25 terms of an AP.whose third term is 7 and seventh term is 2 more than thrice it's third term.

Answer 1

Let a be the first term and d be common difference of given ap.

Given, third term of an ap = 7

⇒ a3 = a + (3 - 1)d

⇒ a + 2d = 7 .... (1)

Again, 7th term is two more than thrice of its third term of given ap.

⇒ a7 = 2 +3a3

⇒ a7 = 2 + 3 × 7 = 2 + 21

⇒ a7 = 23

⇒ a + 6d = 23 .... (2)

On subtracting (1) from (2), we get

a + 6d - (a + 2d) = 23 - 7

⇒ 4d = 16

⇒ d = 4

On putting value of d in (1), we get

a + 8 = 7

⇒ a = 7 -8 = -1

Now, sum of 25 terms of an ap
25/2 (2 (-1)+25-1(4)
25/2×94
1175


hope it help uu

Answer 2

S25th=?
a3rd=7
a7th=2
a+2d=7
a=7-2d
a+6d=2
7-2d+6d=2
4d=-5
d=-5/4
a=7-2d
a=7-2×-5/4
a=7+5/2
a=(14+5)/2
a=19/2
S25th=25/2{2a+(n-1)d}
=25/2(2×19/2+25×-5/4)
=25/2(19-125/4)
=25/2{(76-125)/4}
=25/2(49/4)
=25/2×49/4
=1225/8
=153.125