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Exercise 5.1 · Q14

Q.Find the sum of first 24 terms of the A.P. a1,a2,a3,…a_1, a_2, a_3, \ldots, if it is known that a1+a5+a10+a15+a20+a24=225a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{24} = 225.

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Pairing terms equidistant from the two ends of the 24-term A.P. shows each pair sums to 2a+23d2a+23d; this gives S24=900S_{24}=900.

nnth term: an=a+(n−1)da_n=a+(n-1)d. Sum of nn terms: Sn=n2[2a+(n−1)d]S_n=\dfrac n2\big[2a+(n-1)d\big]. For a set of 2424 terms, terms symmetric about the middle (like a1,a24a_1,a_{24} or a5,a20a_5,a_{20}) each sum to 2a+23d2a+23d.

  1. Note 1+24=251+24=25, 5+20=255+20=25, 10+15=2510+15=25 — each pair of subscripts sums to 2525, i.e. these pairs are symmetric about the centre of the 24-term sequence.
  2. a1+a24=(a)+(a+23d)=2a+23da_1+a_{24}=(a)+(a+23d)=2a+23d.
  3. a5+a20=(a+4d)+(a+19d)=2a+23da_5+a_{20}=(a+4d)+(a+19d)=2a+23d.
  4. a10+a15=(a+9d)+(a+14d)=2a+23da_{10}+a_{15}=(a+9d)+(a+14d)=2a+23d. …

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