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 the 7th term of an AP is 32 and its 13th term is 62. find the AP?

Given

7th term = 32

a+6d=32 ...(1)

13th term = 62

a+12d=62 ...(2)

(2)-(1)

6d = 30

d=5 

(1)==> a+ 30= 32==>a=2

So, a=2 and d=5

So, AP is

2, 7, 12, 17,....

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 A7 = A + 6d = 32 - 1)

and A13 = A + 12d = 62 - 2)

subtracting 1) from 2) :

[A+12d] - [A+6d] = 62-32

or 6d = 30

or d = 5

therefore A + 6d = A + 30 = 32

therefore A = 2

A2 = 7

A3 = 12

and so on

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7th term is a+6d & 13th term is a+12d ,therefore a+12d - a+6d= 62-32=30.6d=30,d=5,so from given info we get a+6d = 32,a=2

therefore A.P=2,7,12,...,n terms

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A7 = A + 6d = 32 - 1)

and A13 = A + 12d = 62 - 2)

subtracting 1) from 2) :

[A+12d] - [A+6d] = 62-32

or 6d = 30

or d = 5

therefore A + 6d = A + 30 = 32

therefore A = 2

A2 = 7

A3 = 12

and so on

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