Question

Find the sum of first 20 terms of an AP in which 3rd term is 7 and 7th term is two more than thrice of its 3rd term.

Answer 1

a+2d=7
a+6d=3(a+2d)+2
a+6d=3*7+2
a7 = 21+2
a7 = 23

a+2d=7
a+6d=23
--------------
-4d = -16
d = 4

a = -1

S20 = n/2[2a+(n-1)d]
= 10 [2(-1) +19*4]
= 10 [-2+76]
= 10[74]
= 740

Answer 2

Answer:

740

Step-by-step explanation:

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

Substitute n = 3

$a_3=a+(3-1)d$

$a_3=a+2d$

we are given that 3rd term is 7

So,$7=a+2d$   --1

we are also given that  7th term is two more than thrice of its 3rd term.

$a_7=3(7)+2$

$a_7=23$

Substitute n = 7 in A

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

$23=a+6d$   --2

Solve 1 and 2

Subtract 1 from 2

$23-7=a+6d-a-2d$  

$16=4d$  

$4=d$  

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

$7=a+2(4)$

$7=a+8$

$-1=a$

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

Now we are supposed to find the sum of first 20 terms :

$S_{20}=\frac{20}{2}(2(-1)+(20-1)4)$

$S_{20}=10(-2+76)$

$S_{20}=740$

Hence the sum of first 20 terms is 740.