Question

An A.P consists of 55 terms of which fifth term is 14 and last term is 114. find the 29th term​

Answer 1

Answer:

STEP – BY – STEP EXPLANATION :

n = 55

a⁵ = 14

a⁵⁵ = 114

a²⁹ = ?

a⁵ = 14 = a + 4d

a⁵⁵ = 114 = a + 54d

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- 100 = - 50d

- 100 / - 50 = d ( minus minus cancel )

2 = d

a⁵ = a + ( n-1 ) d

14 = a + ( 2 ) 2

14 - 4 = a

10 = a

a²⁹ = a + ( n-1 )

= 10 + ( 29 - 1 ) 2

= 10 + 56

= 66..

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Answer 2

Answer:

a29=62

Step-by-step explanation:

Let first term be a and common difference be d

A.T.Q.

a5=a+(5-1)d a55=a+(55-1)d

14=a+4d------ eq 1 114=a+54d----- eq 2

Subtracting eq 1 from eq 2,

114-14=a+54d-a-4d

100=50d

100/50=d

d=2

Putting value of d in eq 1,

a+4×2=14

a=14-8

a=6

For finding 29th term we have,

a=6

d=2

n=29

an=a+(n-1)d

a29=6+(29-1)2

a29=6+28×2

a29=6+56

a29=62