Math, asked by haseebur1414, 10 months ago

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

Answers

Answered by Anonymous
5

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.

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