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Question

The 4th term of an A.P. is 11 and the 8th term exceeds twice the 4th term by 5. Find the A.P. and the sum of first 50 terms.

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Solution

4th term,T4= a+3d = 11

It is given that,
8th term,T8 = a+ 7d= (2×T4 ) +5

So,
T8 = 2×11 +5 = 27

Therefore

a+3d = 11 ---->(1)
a+7d=27----->(2)
Solving
(2) - (1)

4d = 16

Therefore d = 16/4 =5

So, from(1)

a +3 ×4 = 11

So , a = 11-12 = -1

So AP is

-1,3,7,11,15,19,23,27 ......

Sumof 50 terms =

(n/2)×(2a +(n-1)×d)

=(50/2)×(2×-1 +(50-1)×4)

=25×(-2 +49×4) =25×194 = 4656

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