Question

an AP 5,8,11 ... have 40 terms. find the last term . ans the sum of last 10 terms​

Answer 1

Answer:

Last Term = 122

Sum of Last 10 terms = 1085

Step-by-step explanation:

a= 5

d= 3

n= 40

Thus,

      an = a + (n-1)d

           = 5 + (39) 3

            5 +117

           = 122

The last term of the AP is 122

For the sum of last terms,

Let A = 122  D = -3 n = 10

Thus

a10 = 122 + 9(-3)

      = 122 -27

      = 95 = L

Therefore sum of last 10 terms;

n/2 (A + L )  = 5 (122 + 95)

= 5 (217)

= 1085

Answer 2

Given,

An A.P 5, 8, 11, ... till 40 terms.

To Find,

The last term and the sum of the last 40 terms.

Solution,

In the given A.P

d = 8-5 = 3, a = 5

So,

a₄₀ = a+39d

a₄₀ = 5+39*3

a₄₀ = 5+117 = 122

Also,

we will find a₃₁ to get the sum of the last 10 terms.
a₃₁ = a+30d = 5+30*3 = 5+90 = 95

Sum of last 10 terms = 10/2(95+122)

Sum of last 10 terms = 1085

Hence, the last term of the A.P is 122 and the sum of the last 10 terms of the A.P is 1085.