Question

please solve as soon as,..​

Answer 1

ANSWER:-

Given:

In an A.P. of 50 terms,the sum first 10 terms is 210 & the sum of its last 15 terms is 2565.

To find:

Find the A.P.

Solution:

Let a be the first term & d be the common difference of the given A.P.

We know that sum of first term of an A.P is given by;

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

Put n= 10, we get;

${}^{s}10 = \frac{10}{2}[2a + (10 - 1)d] \\ \\ = >210 = 5(2a + 9d) \\ \\ = > 2a + 9d = 42............(1)$

Sum of last 15 terms= 2565

=) Sum of first 50 terms - sum of first 35 terms =2565

=) S50 - S35 = 2565

$= > {}^{s} 50 - {}^{s} 35 = 2565 \\ = > \frac{50}{2} [2a + 50 - 1) d] - \frac{35}{2} [2a + (35 - 1)d]= 2565 \\ \\ = > 25(2a + 49d) - \frac{7}{2} (2a + 34d) = 513 \\ \\ = >5(2a + 49d) - \frac{7}{2} (2a + 34 d) =513 \\ \\ = > 3a + 126d = 513 \\ \\ = > a + 42d = 171..............(2)$

Multiply (2) with 2 we get;

=) 2a +84d = 342............(3)

Now,

Subtracting (1) from (3), we get;

=) 84d -9d = 342 -42

=) 75d= 300

=) d= 300/75

=) d= 4

So,

From equation (2), we get;

=) a +42(4) = 171

=) a + 168 = 171

=) a= 171 -168

=) a= 3

Therefore,

The A.P. is a , a + d, a+2d , a+3d

So,

A.P. is 3, 7, 11, 15.....

Hope it helps ☺️