Question

Find the A.P whose tenth term 52 and whose 17th term exceeds its 13th term by 20

Answer 1

The AP is shown in the picture

Answer 2

The required A.P is 7,12,17,22,.....

Step-by-step explanation:

To find : The A.P whose tenth term 52 and whose 17th term exceeds its 13th term by 20 ?

Solution :

The nth term of A.P is $a_n=a+(n-1)d$.

17th term exceeds its 13th term by 20 ,

i.e. $(a+16d)=(a+12d)+20$

$16d-12d=20$

$4d=20$

$d=\frac{20}{4}$

$d=5$

The tenth term of A.P is 52.

i.e. $a+9d=52$

Substitute d=5,

$a+9(5)=52$

$a+45=52$

$a=52-45$

$a=7$

The first term is 7 and the common difference is 5.

Second term is $a+d=7+5=12$

Third term is $a+2d=7+2(5)=17$

Fourth term is $a+3d=7+3(5)=22$

and so on..

The required A.P is 7,12,17,22,.....

#Learn more

If the sum of first 7 terms of an ap is 119 and the sum of first 17 term is 629 ind the sum of first n term

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