Question

An AP 5, 8, 11…has 40 terms. Find the last term. Also find the sum of the last 10 terms.

pls answer with explanation​

Answer 1

Step-by-step explanation:

a= 5, d= 3, n =40
Using formula An = a+ (n-1)d,
An= 5+ (39)3
An=117+5
An= 122
∴ The last term of this AP is 122

Reversing the AP, a= 122, d=-3, n=10
Using formula, Sn == (n(2a+(n-1)d))/2
Sn= (10(2*122+ (9)(-3)))/2
Sn= (5(244+ (-27)))
Sn = 5(217)= 1085
∴ The sum of last 10terms of this AP is 1085

Answer 2

Answer:

1085

Step-by-step explanation:

first Term of the AP (a) = 5

Common difference (d) = 8 - 5 = 3

Last term = a40 = a + (40 - 1) d

= 5 + 39 × 3 = 122

Also a31 = a + 30d = 5 + 30 × 3 = 95

Sum of last 10 terms =  n 2 n2(a31 + a40)

=  10 2 102(95 + 122)

= 5 × 217 = 1085