if 24th term of ap is twice of its 10 th term show that 72nd term is 4 times of 15 the term
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Given: a24= 2(a10)
ie: a+23d= 2(a+9d)
Rearrange the above equation as:
2a-a = 23d-18d
ie: a= 5d ----(i)
a72= a+71d= 5d+71d {by (i)
ie: a72= 76d----(ii)
a15= a+14d= 5d +14d {by (i)
ie: a15= 19d ----(iii)
take a72=76d {by (ii)
a72=4 × 19d
a72= 4× a15
Hence proved
ie: a+23d= 2(a+9d)
Rearrange the above equation as:
2a-a = 23d-18d
ie: a= 5d ----(i)
a72= a+71d= 5d+71d {by (i)
ie: a72= 76d----(ii)
a15= a+14d= 5d +14d {by (i)
ie: a15= 19d ----(iii)
take a72=76d {by (ii)
a72=4 × 19d
a72= 4× a15
Hence proved
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