Question

the 5th and the 12th term of an A.P are 23 and 37 respectively. find the sum of the first 40 terms of the A.P​

Answer 1

We know that nth term in an AP is given by - A_n=a+(n-1)d; where a is the first term and d is common difference

According to the question-

a+4d=23...(i)

a+11d=37...(ii)

Solving (i) and (ii) simultaneously, we get -

a=15

d=2

For sum upto 40 terms,

S=2a+(n-1)d

S=2*15 + 39*2

S= 30+78

S = 108

Hope this helped :)

Answer 2

Answer: a5=23

a12=37

a5=a+4d=23...(1)

a12=a+11d=37...(2)

By elimination method,

-7d=-14

d=-14/-7=2

Put d=2in equation 1

a+4×2=23

a=23-8=15.

an=a+(n-1)d

a40=15+(40-1)2

a40=15+39×2

a40=15+78=93

Sn=n/2(a+an)

S40=40/2(15+93)

S40=20×108=2160

Hence the sum of first 40 terms are 2160

I hope this is sufficient answer

Step-by-step explanation: