Question

7 term of an Ap is 40 then the sum of frist 13 term is

Answer 1

Hey there,

a7 = 40
a + 6d = 40
a = 40 - 6d

Sn = 13/2[2a + (13-1)d]
= 13/2[2(40-6d) + 12d]
= 13/2[80-12d + 12d]
= 13/2[80]
= 13 x 40
= 520

So, sum of first 13 term is 520

Hope it helps!

Answer 2

Hi !

a₇ = 40
i.e

a + 6d = 40        
a = 40 - 6d              ----> (1)

-------------------------------------------

Sn = n/2 [ 2a + (n-1)d]
     = 13/2 [ 2a + 12d ]
     = 13/2 x 2 [a + 6d]
     = 13 [a + 6d]
     = 13a + 78d

S₁₃ = 13a + 78d

From (1) ,

13 ( 40 - 6d) + 78d = S₁₃

520 - 78d + 78d = S₁₃

S₁₃ = 520

∴ Sum of first 13 terms = S₁₃ = 520