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Question

# Find the sum of the following arithmetic progressions: (i) 50, 46, 42, ...... to 10 terms (ii) 1, 3, 5, 7, ..... to 12 terms (iii) 3,92,6,152,……25 terms (iv) 41, 36, 31, ..... to 12 terms (v) a + b, a - b, a - 3b, ... to 22 terms (vi) (x−y)2,(x2+y2),(x+y)2,……to n terms (vii) x−yx+y,3x−2yx+y,5x−3yx+y,……to n terms

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Solution

## (i) 50, 46, 42, ..... 10 terms a = 50, d = (46 - 50) = - 4 n = 10 Sn=n2[2a+(n−1)d]S10=102[2×50+(10−1)(−4)]=320 (ii) 1, 3, 5, 7, ..... 12 terms S12=122[2×1+(12−1)2]=6×24=144 (iii) 3,92,6,152,……25 termsSn=n2[2a+(n−1)d]S25=252(2×3+24×32)=525 (iv) 41,36,31,……12 termsSn=n2[2a+(n−1)d]S25=122[2×41+(11)(−5)]=162 (v) a+b,a−b,a−3b,……to 22 termsSn=n2[2a+(n−1)d]S22=222[2a+2b+21(−2b)]=22a−440b (vi) (x−y)2,(x2+y2),(x+y)2,……n termsSn=n2[2a+(n−1)d]=n2[2(x2+y2−2xy)+(x−1)(−2xy)]=n[(x−y)2+(n−1)xy] (vii) x−yx+y,3x−2yx+y,5x−3yx+y,……to n terms nth term in above sequence is (2n−1)x−nyx+y Sum of n terms is given by 1x+y[x+3x+5x+……+(2n−1)x−(y+2y+3y……+ny)]=1x+y[n2(2x+(n−1)2x)−n(n+1)y2]=12(x+y)[2n2x−2n2y−ny]=n2(x+y)[2nx−2ny−y]=n2(x+y)[n(2x−2y)−y]

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