Question

in an A.P 19th term is 52 and 38th term is 128,find sum of first 56 term.​

Answer 1

heya mate ur answer is in attachments to

Answer 2

SOLUTION:-

Given:

In an A.P. 19th term is 52 & 38th term is 128.

To find:

The sum of first 56 term.

Explanation:

We have,

${}^{t} 19 = 52 \: \: and \: \: {}^{t}38 = 128$

Using the formula of the nth term of A.P.

${}^{t} n = a + (n - 1)d$

• a= first term
• d= common difference
• tn= nth term

Therefore,

${}^{t} 19 = a + (19 - 1)d \\ \\ 52 = a + 18d..............(1)$

&

${}^{t} 38 = a + (38 - 1)d \\ \\ 128 = a + 37d..................(2)$

Subtracting equation (1) from equation (2), we get;

=) 128 -52 = (a-a) + (37d -18d)

=) 76 = 19d

=) d= 76/19

=) d= 4

Substituting the value of d in equation (1), we get;

=) 52= a+ 18 × 4

=) 52 = a + 72

=) a= 52 -72

=) a= -20

Now,

Find the value of S56 we will using the formula of sum of n term;

${}^{s} n = \frac{n}{2} (2a + (n - 1)d)$

Therefore,

Substituting the given value in formula we can find the value of Sn.

${}^{S} 56 = \frac{56}{2} (2 \times ( - 20) + (56 - 1) \times 4) \\ \\ {}^{S} 56 = 28 \times ( - 40 + 55 \times 4) \\ \\ {}^{S} 56 = 28 \times ( - 40 + 220) \\ \\ {}^{S} 56 = 28 \times 180 \\ \\ {}^{S} 56 = 5040$

Thus,

The sum of first 56 term is 5040.