Question

find the sum orithmetic progression
1/15,1/12,1/10,........up to 11 terms​

Answer 1

a=1\15

d=1\12-1\15

=-1\3

n=11

Sn=n\2(2a+(n-1)d)

=11\2(2(1\15)+(11-1)-1\3

= 11\2(2\15)+(10)-1\3

= 11\15 -0.3

=0.4333333

I hope this will help u

Answer 2

Answer:

The sum of 11 terms in the given arithmetic progression is $\dfrac{99}{60}$.

Step-by-step explanation:

Arithmetic progression is a sequence of terms in which difference between two terms is a constant.

Arithmetic sequence is given by

$a, a + d, a + 2d, ......, a + (n-1)d$

Generalizing the sequence, the nth term is given $a_n= a_1 + (n-1)d$

where $a_1$ is the first term, $d$ is the common difference and $n$ is the

number of terms.

And sum of n terms in arithmetic progression is given by

$S_n=\dfrac{n}{2} \big (2a+(n-1)d\big)$

Given the series $\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10},........$ and $n=11$.The first term is $a=\dfrac{1}{15}$ The difference between the two numbers is $d=\dfrac{1}{12}-\dfrac{1}{15}=\dfrac{5-4}{60}=\dfrac{1}{60}$Substituting the values, $S_11=\dfrac{11}{2} \Bigg (2\times \dfrac{1}{15} +(11-1) \dfrac{1}{60}\Bigg)$$=\dfrac{11}{2} \Bigg ( \dfrac{2}{15} + \dfrac{10}{60}\Bigg)=\dfrac{11}{2} \times \dfrac{8+10}{60}$$=\dfrac{11}{2} \times \dfrac{18}{60}=\dfrac{99}{60}$Therefore, the sum of 11 terms in the given arithmetic progression is $\dfrac{99}{60}$.