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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) (xy)2,(x2+y2),(x+y)2,to n terms
(vii) xyx+y,3x2yx+y,5x3yx+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+(n1)d]S10=102[2×50+(101)(4)]=320

(ii) 1, 3, 5, 7, ..... 12 terms
S12=122[2×1+(121)2]=6×24=144

(iii) 3,92,6,152,25 termsSn=n2[2a+(n1)d]S25=252(2×3+24×32)=525

(iv) 41,36,31,12 termsSn=n2[2a+(n1)d]S25=122[2×41+(11)(5)]=162

(v) a+b,ab,a3b,to 22 termsSn=n2[2a+(n1)d]S22=222[2a+2b+21(2b)]=22a440b

(vi) (xy)2,(x2+y2),(x+y)2,n termsSn=n2[2a+(n1)d]=n2[2(x2+y22xy)+(x1)(2xy)]=n[(xy)2+(n1)xy]

(vii) xyx+y,3x2yx+y,5x3yx+y,to n terms
nth term in above sequence is (2n1)xnyx+y
Sum of n terms is given by
1x+y[x+3x+5x++(2n1)x(y+2y+3y+ny)]=1x+y[n2(2x+(n1)2x)n(n+1)y2]=12(x+y)[2n2x2n2yny]=n2(x+y)[2nx2nyy]=n2(x+y)[n(2x2y)y]


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