Question

In an AP, a=8,Tn=33,Sn=123, find d and n.

Answer 1

Given: a= 8

t(n) = 33

a+(n-1)d = 33

8+(n-1)d = 33

(n-1)d =25.......(1)

Also,
S(n) = 123

n/2 [2a+(n-1)d] = 246

n [ 2(8)+25 ] =246

n [41] =246

n = 6

(1) ==>

5d =25

d = 5

Answer 2

Hi ,

It is given that ,

In A.P , first term = a = 8 ,

Let common difference = d

nth term = Tn = 33

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If a , d are first term and common

difference of an A.P

nth term = Tn = a + ( n - 1 )d

Sum of n terms = Sn = n/2 [ a + Tn ]

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1 ) Sn = 123

n/2 [ 8 + 33 ] = 123

( n/2 ) × 41 = 123

n = ( 123 × 2 )/41

n = 6

2 ) Tn = 33

a + ( n - 1 )d = 33

8 + ( 6 - 1 )d = 33

5d = 25

d = 25/5

d = 5

Therefore ,

d = 5 , n = 6

I hope this helps you.

: )