Question

How many terms of A.P.78,71,64............are needed to give the sum 465? Also find the last term of this ap

Answer 1

Answer:-

n = 10

l = 15

Given :-

$s_n = 465$t

a = 78

d = 71 - 78

d = -7

To find :-

The number of term need to make the sum 465.

The last term of A.P

Solution:-

Let the nth term of AP is needed to make sum 465.

By using sum formula :-

$\huge \boxed {s_n = \dfrac{n}{2}[2a + (n-1) d]}$

Now, put the given value,

→$465 = \dfrac{n}{2}[2\times 78 + (n-1) -7 ]$

→$465 \times 2 = n ( 156 -7n +7)$

→$930 = n ( 163 -7n )$

→$930 = 163n - 7n^2$

→$7n^2 -163n +930 = 0$

→$7n^2 -70n -93n +930 = 0$

→$7n ( n - 10) -70 ( n - 10) = 0$

→$(7n -70) (n-10) = 0$

→$7n -70 = 0 , n-10 = 0$

→$n = \dfrac{70}{7} , n = 10$

→$n = 10 , n = 10$

hence,

The 10 th term will give the sum 465.

Now,

Last term of A. P is given by :-

→$a_l = a + (n-1) d$

→$a_l = 78 + (10-1) -7$

→$a_l = 78 + 9 \times -7$

→$a_l = 78 -63$

→$a_l = 15$

hence,

The last term of A. P will be 15.