Question

find the sum of 1 /15,1/12,1/10.......to 11terms​

Answer 1

$\huge\mathscr\pink{Hey \: Mate}$

See attachment for your answer

Hope it will help you ☺️

Answer 2

Given :

$\sf a = \frac{1}{15} \\ \\ \sf d = \frac{1}{12} - \frac{1}{15} = \frac{1}{60} \\ \\ \sf n = 11$

To Find :

• Sum of 11 terms

Solution :

$\large \boxed{ \bf \green{S_n = \frac{n}{2} \bigg[2a + (n - 1)d \bigg]} }\\ \\ \implies \sf \frac{11}{2} \bigg[ 2 \times \frac{1}{15} +(11 - 1) \times \frac{1}{60} \bigg] \\ \\ \implies \sf \frac{11}{2} \times \bigg[ \frac{2}{15} + \frac{\cancel{10}}{\cancel{60}}\bigg] \\ \\ \implies \sf \frac{11}{2} \times \bigg[ \frac{2}{15} + \frac{1}{6}\bigg] \\ \\ \implies \sf \frac{11}{2} \times \bigg[ \frac{4+5}{30} \bigg]\\ \\ \implies \sf \frac{11}{2} \times \frac{\cancel{9}}{\cancel{30}} \\ \\ \implies \sf \frac{11}{2} \times \frac{3}{10} \\ \\ \large \boxed{ \bf \blue{S_n =\frac{33}{20} }}$

$\large \underline{ \bf Sum \: of \: given \: terms \: is \: \frac{33}{ 20} }$