Question

How Many terms of Ap 78,71,64 are needed to give the sum 465 also find the last term of Ap

Answer 1

a= 78 
d= 71-78= -7
Sn= 465

Sn= n/2 (2a +(n-1)d)
465= n/2 (2 x 78 + (n-1) -7)
930= n (156 +7 -7n)
930 = n(163 - 7n)
7n² - 163n +930= 0
 by making factors we will get
(7n - 93) (n-10)

sice n can not be a fraction so n=10
an= a + (n-1)d
= 78 - 7(10-1)
= 78 - 7(9)
=78 - 63 
= 15

Answer 2

Answer:

15

Step-by-step explanation:

a= 78 

d= 71-78= -7

Sn= 465

Sn= n/2 (2a +(n-1)d)

465= n/2 (2 x 78 + (n-1) -7)

930= n (156 +7 -7n)

930 = n(163 - 7n)

7n² - 163n +930= 0

 by making factors we will get

(7n - 93) (n-10)

sice n can not be a fraction so n=10

an= a + (n-1)d

= 78 - 7(10-1)

= 78 - 7(9)

=78 - 63 

= 15