Question

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

Answer 1

Answer:

Sum of the first n terms is given by

S

n

=

2

n

[2a+(n−1)d]

Putting n = 10, we get

S

10

=

2

10

[2a+(10−1)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

S

50

−S

35

=2565

2

50

[2a+(50−1)d]−

2

35

[2a+(35−1)d]=2565

25(2a+49d)−

2

35

(2a+34d)=2565

5(2a+49d)−

2

7

(2a+34d)=513

10a+245d−7a+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=42−36

2a=6

a=3

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