Question

An Arithmetic sequence has its 5th term equal to 22 and its 15th term equal to 62 find its 100th term and sum of first 50 terms

Answer 1

5th term
a + 4d = 22
a= 22 - 4d
15th term
a+ 14d = 62
22-4d +14d = 62
d= 40/10
d= 4
using first equation...
a+4d = 22
a+ 16 = 22
a= 6
100th term
a+ 99d =?
6 + 396
= 402
sum of first 50 terms
x =(n/2)[2a +({n-1}d]
= (25)( 12 +(49 X 4))
5200    (calculate)

Answer 2

Answer :-

Formula used : aₙ = a + (n - 1)d

Case I :

An arithmetic sequence has it's 5th term to equal 22.

=> 22 = a + (5 - 1)d

=> a = 22 - 4d _____ (i)

Case II :

it's 15th term equal 62.

=> 62 = a + (15 - 1)d

=> 62 = 22 - 4d + 14d [from equation (i)]

=> 10d = 40

=> d = 40/10 = 4

Putting the value of d in equation (i) we get,

=> a = 22 - 4 × 4

=> a = 22 - 16 = 6

Now,

It's 102nd term = 6 + (102 - 1)4

= 6 + (101)4

= 6 + 404 = 410

Hence,

It's 102nd term = 410