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Standard X

Mathematics

Arithmetic Progression

The 4th term ...

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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