Question

The sum of the series in A. P. is 40*1/2, the common difference is 2*1/2, the common difference is 25 and the last term is 13. answer it ​

Answer 1

Answer:

hope u understood

Step-by-step explanation:

n/2 (t1+tn) = Sn

n/2 (t1 + 13) = 41.5 (which is 81/2)

n/2 (t1 + 13) = 81/2

n (t1 + 13) = 81 -(1)

Then,

t1 + (n-1)d = tn

t1 + (n-1)5/2 = 13 -(2)

From equation 1,

n = 81/(t1+13) -(3)

Substituting (3) in (2)

t1 + {81/(t1+13) - 1}5/2 = 13

Solving this we will get,

2t1(t1+13)+5(68−t1)=13×2×(t1+13)

2(t1)2+26t1+340−5t1=26t1+338

2(t1)2+340−5t1=338

2(t1)2−5t1+2=0

t1 = 2 or 0.5

We got two values for t1. Substituting them in (3) for finding n

If t1 = 2, n = 81/(2+13) = 5.4

So, t1 = 2 is not possible

if t1 = 0.5, n == 81/(0.5+13) = 6

Thus we get first term as 0.5 and number of terms as 6.

The series is 0.5, 3, 5.5, 8, 10.5, 13

Thank u