Latest TCS Aptitude Question SOLUTION: sum of 100 terms common to progressions 17,21,25,29..... and 116,21,26,......

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sum of 100 terms common to progressions 17,21,25,29..... and 116,21,26,......

Asked In TCS SHUBHAM SONI (9 years ago)

Unsolved Read Solution (6)

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Sayandip Dey  (9 years ago)

first term 21;
common diff=20;
sum=100/2(2*21+99*20)=101100

Is this solution Helpfull? Yes (15) | No   

Simranjit Singh  (9 years ago)

Terms common to both the expressions are:

21,41,61,81.....

Here First term(a)=21, Common Difference(d)=20, n=100

So sum of terms in the AP = (n/2)*(a+(n-1)*d)
= (100/2)*(21+(100-1)*20)
= 50*(21+99*20)
= 50*2001
= 100050

Is this solution Helpfull? Yes (2) | No (10)  

indrajeet singh  (9 years ago)

ans is....4220

common term is 21,41,61......401
so total term is 20 and d=20
so sum is 20/2(21+401)=4220

Is this solution Helpfull? Yes (2) | No (5)  

Balaji  (9 years ago)

average of first and last term = (17 + (17 +(100*4)/2 = (17+417)/2 = 217
sum of 100 terms = 217 * 100 = 21700

Is this solution Helpfull? Yes | No (2)  

Ritesh Kumar  (9 years ago)

17 21 25....
21 26 31...
common terms of both series r 21 41 61 so in AP wit cd 20
100th term is 2022
therefore sum is 100/2(21+2022)=10100

Is this solution Helpfull? Yes | No (5)  

anand jha  (9 years ago)

first term is 21 then 41,61 upto 100 term the 100 term is 2001.
therefore sum is..
100/2{21+2001}

2022*50= 101100 ans.

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