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If A1,A2,A3,....is in AP such that, A1+A5+A10+A15+A20+A24=225, then, A1+A2+A3+...A23+A24 is equal to?
A1,A2,A3,....is in APthe write in common terms of a and d..i.e.A1 = aA2 =a+dA3 =a+2d..A24 = a+23dsumming all we get, 4(6a+69d) = > 4*225 = 900 perhaps.
u can also solve in such way,(A1+A24) +(A5+A20) +(A10+A15)=225.THEREFORE 3(A1+A24)=225(A1+A24)=75.NOW SUM=24/12(A1+A24)=900
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