Question

Find the 20th term of the ap whose third term is 7 and the 7th term exceeds three times the third term by 2 also find its nth term

Answer 1

Answer:

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Answer 2

The 20th term of the A.P. is 58 .

The nth term is $a_n=3n-2$  .

Explanation:

The n th term of AP = $a_n=a+(n-1)d$           (1)

, where a = first term and d is common difference.

Given : The third term of an AP is 7.

i.e. $a_3=a+2d=7$              (2)

7th term = $a_7=a+6d$

if 7th term exceeds three times the third term by 2  , then we have

$3a_3-a_7=2$

$3(7)-(a+6d)=2$

$21-(a+6d)=2$

$a+6d=19$              (3)

Subtract (2) from (3) , we get

$4d=12\Rightarrow\ d=3$

Put d=3 in (2) ,we get

$a+2(3)=7$

$a+6=7\Rightarrow\ a=1$

Now , put value of a and d in (1), the nth term will be

$a_n=1+(n-1)3=1+3n-3=3n-2$

i.e. $a_n=3n-2$

At n= 20 , we get

$a_{20}=3(20)-2=60-2=58$

So , the 20th term of the A.P. is 58 .

#Learn more :

The third term of an ap is 8 and the ninth term of an ap exceeds three times the third term by 2 find the sum of its first 19 terms​

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