The sum of the first 10 terms of an A.P is 160 and the sum of the next 10terms of the the same A.P is 360 .Find the common diffrence of the A.P
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Let the AP be a , a+d , a+2d , a+3d......
thus 1st term of the AP=a
10th term of the AP=a+9d
Sn=n/2[a+l]
S10=10/2[a+a+9d]
160=5[2a+9d]
160=10a+45d....... i)
11th term =a+10d
20th term=a+19d
S10=10/2[a+10d+a+19d]
360=5[2a+29d]
360= 10a+145d.......ii)
on solving i & ii
d=2
thus 1st term of the AP=a
10th term of the AP=a+9d
Sn=n/2[a+l]
S10=10/2[a+a+9d]
160=5[2a+9d]
160=10a+45d....... i)
11th term =a+10d
20th term=a+19d
S10=10/2[a+10d+a+19d]
360=5[2a+29d]
360= 10a+145d.......ii)
on solving i & ii
d=2
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