Question

If the 8th term of an A.P is 37 and the 15th term is 15 more than the 12th term,find the A.P. Also,find the sum of first 20 terms of this A.P

Answer 1

Given ;
t8 term =a+7d=37 --- equation (1)

To find :
A.p and sum of first 20 terms in A.P

solution :

Given that  15th term is 15 more than the 12th term,

a+14d = 15 + a+11d
14d-11d =15

3d=15

d=5

Substitute the value of d in t8 term we get

a+7(5)=37

a+35=37

a=2

∴a=2 d=5

Now A.P terms will be:
 First term a = 2

Second term : a +d=  7

Third term - a+2d = 12

Fouth term= a+3d = 17 and so on

∴ A.P is 2,7,12,17......

Sum of first 20 terms in A.p=sn=n/2[2a+(n-1)d]

=20/2[2x2+(20-1)x5]

=10[4+19x5]

=10[4+95]

=10x99= 990

∴Sum of first 20 terms in A.P is 990

Answer 2

Let a = first term , d = common difference

eighth term = a8 = a + 7d = 37    ........(1)

Also 15th​ term = 15 + 12th term

⇒a15 = 15 + a12⇒a + 14d = 15 + a + 11d⇒14d − 11d = 15⇒3d = 15⇒d = 5from (1), we get a + 7×5 = 37so, a + 35 = 37a = 2.so the required AP is a, a + d, a + 2d,......... or 2, 7, 12, ..........We know that, Sn = n2[2a + (n−1)d]S15 = 152[2(2) + 14×5]S15 = 15(2 + 35) = 15 × 37 = 555