Question

the sum of 6 terms which form an AP is 345 the different of first & last terms is 55 find the

Answer 1

The general form A.P. is a, a+d, a+2d, a+3d, ......, a+(n-1)d.

But there are only six terms are there so
a, a+d, a+2d, a+3d, a+4d, a+5d.

Sum of six terms is 345.

so a+a+d+a+2d+a+3d+a+4d+a+5d = 345
6a + 15d = 345------------(1)

Differnce between last term and first term is 55.

so

(a+ 5d) – (a) = 55

5d = 55

d = 11 --------------(2)

Now we find the first term of A.P. by substitute in eq 1.

6a + 15d = 345

6a + 15 (11) = 345

6a + 165 = 345

6a = 345 – 165

6a = 180

a = 30

then,

a = 30, d = 11,

so

A.P. is 30, 41,52,63,74,85.


hope it will help you

Devil_king ▄︻̷̿┻̿═━一

Answer 2

heya mate !!

here's your answer :-
________________________

Sum of the six term = 345

The differences between first term and second term is = 55.

First term is = a

And last term = a + 5d

So , According to the question

==> a - a + 5d = 55

==> 5d = 55

==> d = 11
___________________________

General form of an A.P. of 6th term =

==> a + a + d + a + 2d + a + 3d + a + 4d + a + 5d = 345

==> 6a + 15d = 345

On putting the value of " d "

==> 6a + 15 * 11 = 345

==> 6a + 165 = 345

==> 6a = 345 - 165

==> 6a = 180

==> a = 30
__________________________

We get ,

First term = 20

difference = 11
__________________________

So , the arithmetic progress of 6th term =

30 , 41 , 52 , 63 , 74 , 85
__________________________

hope it helps !!

thanks for asking :)

☆ be brainly ☆