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Exercise 5.1 · Q4

Q.If the fourth term of an A.P. is 4, then the sum of its 7 terms is:

(a) 28
(b) 26
(c) 32
(d) none of these
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For an odd number of terms, the sum of an A.P. equals (number of terms) ×\times (middle term); here the 44th term is exactly the middle (4th of 7) term, so the sum of 77 terms is simply 7×a47\times a_4.

Sum of nn terms of an A.P.:

Sn=n2[2a+(n−1)d]=n2(a1+an)S_n = \frac{n}{2}\big[2a+(n-1)d\big] = \frac{n}{2}(a_1+a_n)

  1. Given: 44th term a4=a+3d=4a_4 = a+3d = 4, where aa is the first term and dd the common difference.
  2. Write the sum of 77 terms using the standard formula:

S7=72[2a+(7−1)d]=72[2a+6d]S_7 = \frac{7}{2}\big[2a+(7-1)d\big] = \frac{7}{2}[2a+6d]

  1. Factor out 22 from the bracket:

S7=72×2(a+3d)=7(a+3d)S_7 = \frac{7}{2}\times 2(a+3d) = 7(a+3d)

  1. Substitute a+3d=a4=4a+3d = a_4 = 4 from Step 1: S7=7×4=28S_7 = 7\times 4 = 28 …

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