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8. Arithmetic Progressions (AP)
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Q7 of 176 Page 8

Find the sum of first 25 terms of an A.P. whose nth term is 1 — 4n.

Given: an = 1 – 4n


Taking n = 1,


a1 = 1 – 4(1) = 1 – 4 = -3


Taking n = 2,


a2 = 1 – 4(2) = 1 – 8 = -7


Taking n = 3,


a3 = 1 – 4(3) = 1 – 12 = -11


Therefore the series is -3, -7, -11, …


So, a = -3, d = a2 – a1 = -7 – (-3) = -7 + 3 = -4


Now, we have to find the sum of the first 25 terms of the AP






⇒ S25 = 25 × (-51)


⇒ S25 = -1275


Hence, the sum of 25 terms is -1275.


More from this chapter

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6

If the nth term of an A.P. is (2n + 1), find the sum of first n terms of the A.P.

7

If the nth term of an A.P. is 9 — 5n, find the sum to first 15 terms.

8

If the sum to n terms of a sequence be n2 + 2n, then prove that the sequence is an A.P.

9

Find the sum to first n terms of an A.P. whose kth term is 5k + 1.

Questions · 176
8. Arithmetic Progressions (AP)
1 1 1 1 1 1 2 2 2 2 3 3 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 1 1 1 1 2 3 4 5 5 6 7 7 7 7 7 8 8 9 9 9 10 11 12 13 14 15 16 17 17 17 17 18 18 18 18 18 18 18 18 18 19 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 1 2 3 3 4 4 5 6 7 7 7 8 9 10 10 10 1 2 3 3 4 5 6 7 7 8 9 10 11 12 13 14 15 15 16 17 18 19 20 20 21 22 23 24 24 25 26 27 28 29 30 31 32 33 34 34 35 36 37 38 39 40 41 42 43 44 44 45 46 47 48
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