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Which term of the AP 3,15,27,39,...will be 120 more than its 21st term?

 

 
 
 
 
 

Answers (1)

3,15,27,39,...(given)

We know a_n=a+(n-1)d   (for an AP).....(1)

For a_{21},

a=3

d=12

n=21

a_{21}=3+(21-1)12

a_{21}=3+240

a_{21}=243

Lets assume the term l which is 120 more than a_{21}

So value of l = 243+120=363

Putting the VALUE IN EQUATION (1)

363=3+(n-1)12

363=3+12n-12

363=12n-9

12n=372

n=31

The required number of the term is 31st.

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Safeer PP

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