The sum of n terms of an AP whose first term is 6 and common difference is 40 is equal to the sum of 2n terms of another AP whose first term is 40 and common difference is 6. Find n.
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First A.P.
a=6,d=40
Second A.P.
a=40,d=6
Sn=2Sn
n/2(2a+(n-1)d)=2×n/2(2a+(n-1)d)
2×6+(n-1)×40=2(2×40+(n-1)6)
12+40n-40=2(80+6n-6)
40n-28=160+12n-12
40n-28=148+12n
40n-12n=148+28
28n=176
n=176/28
a=6,d=40
Second A.P.
a=40,d=6
Sn=2Sn
n/2(2a+(n-1)d)=2×n/2(2a+(n-1)d)
2×6+(n-1)×40=2(2×40+(n-1)6)
12+40n-40=2(80+6n-6)
40n-28=160+12n-12
40n-28=148+12n
40n-12n=148+28
28n=176
n=176/28
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