Question

Find the sum of first 25 terms of an
a.p., in which the third term is 7 and seventh term is two more than thrice of its third term

Answer 1

a+2d=7
a+6d=2+3(a+2d)
=2+3a+6d
from above
a=-1,Putting in first eqation
d=4
hence sum of 25 terms =25/2(-2+(24)d)
=25/2(94)
=25×47
=1175

Answer 2

Answer:

1175

Step-by-step explanation:

Formula of nth term = $a_n=a+(n-1)d$

Substitute n = 3

So,  $a_3=a+(3-1)d$

$a_3=a+2d$

Since we are given that  the third term is 7

So,  $a_3=a+2d =7$  -1

Substitute n = 7

$a_7=a+(7-1)d$

$a_7=a+6d$

Since we are given that seventh term is two more than thrice of its third term.

$a_7=3a_3+2$

$a_7=3(7)+2$

$a_7=23$

So, $a_7=a+6d =23$  --2

Subtract 1 from 2

$a+6d-a-2d=23-7$

$4d=16$

$d=4$

Substitute the value of d in 1 to get value of a .

$a+2(4) =7$

$a+8=7$

$a=-1$

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

Substitute n =25

$S_{25}=\frac{25}{2}(2(-1)+(25-1)4)$

$S_{25}=\frac{25}{2}(-2+96)$

$S_{25}=1175$

Hence the sum of first 25 terms is 1175