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You have to use the sum formula Sn=n/2 (2a+(n-1)d) in this case n is m m/2 (2x161+(m-1)x-7)=0 322-7m+7=0 7m=329 m=47 (b) in geometric progression the sum of n terms formula is a(1-r^n)/(1-r) and sum to infinity formula is a/1-r s0 this becomes a(1-r^n)/(1-r)<0.9 a/1-r as...

salam can someone help with question 1 http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w11_qp_13.pdf