Question

the sum of first nth term of the series a,3a,5a i​

Answer 1

Answer:

AP is a,3a,5a,....

AP is a,3a,5a,....Here, a=a, (first term)

AP is a,3a,5a,....Here, a=a, (first term)d=2a (common difference)

AP is a,3a,5a,....Here, a=a, (first term)d=2a (common difference)Sum =S

AP is a,3a,5a,....Here, a=a, (first term)d=2a (common difference)Sum =S n

AP is a,3a,5a,....Here, a=a, (first term)d=2a (common difference)Sum =S n

AP is a,3a,5a,....Here, a=a, (first term)d=2a (common difference)Sum =S n =n/2[2a+(n−1)d] , where n= number of terms

AP is a,3a,5a,....Here, a=a, (first term)d=2a (common difference)Sum =S n =n/2[2a+(n−1)d] , where n= number of terms=n/2[2a+2an−2a]

AP is a,3a,5a,....Here, a=a, (first term)d=2a (common difference)Sum =S n =n/2[2a+(n−1)d] , where n= number of terms=n/2[2a+2an−2a]=an

AP is a,3a,5a,....Here, a=a, (first term)d=2a (common difference)Sum =S n =n/2[2a+(n−1)d] , where n= number of terms=n/2[2a+2an−2a]=an 2

AP is a,3a,5a,....Here, a=a, (first term)d=2a (common difference)Sum =S n =n/2[2a+(n−1)d] , where n= number of terms=n/2[2a+2an−2a]=an 2 .

Answer 2

Answer:

Sn= a+(n-1)*2

Sn= a+2n-2