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Question

In an AP of 50 term the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565 Find the AP.

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Solution

Sum of the first n terms is given by

Sn=n2[2a+(n1)d]

Putting n = 10, we get

S10=102[2a+(101)d]

210=5(2a+9d)

2a+9d=42 ...............(1)

Sum of the last 15 terms is 2565

Sum of the first 50 terms - sum of the first 35 terms = 2565

S50S35=2565

502[2a+(501)d]352[2a+(351)d]=2565

25(2a+49d)352(2a+34d)=2565

5(2a+49d)72(2a+34d)=513

10a+245d7a+119d=513

3a+126d=513

a+42d=171 ........(2)

Multiply the equation (2) with 2, we get

2a+84d=342 .........(3)

Subtracting (1) from (3)

d=4

Now, substituting the value of d in equation (1)

2a+9d=42

2a+9×4=42

2a=4236

2a=6

a=3

So, the required AP is 3,7,11,15,19,23,27,31,35,39........

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