Question

the sum of 11 terms of an A.P whose 6th term is 5 is_ _ _ _ _​

Answer 1

Answer:

Given,

\begin{gathered}{ \star{ \mathbb{n = 6}}} \\ { \star{ \mathbb{a _{n} = 5}}}\end{gathered}

⋆n=6

⋆a

n

=5

\begin{gathered}a_{n} = a + (n - 1)d \\ 5 = a + (6 - 1)d \\ 5 = a + 5d \\ a = 5 - 5d\end{gathered}

a

n

=a+(n−1)d

5=a+(6−1)d

5=a+5d

a=5−5d

Now to find the sum of 11 terms of the A.P

given,

\begin{gathered}{ \star{ \mathbb{n = 11}}} \\ { \star{ \mathbb{a = 5 - 5d}}}\end{gathered}

⋆n=11

⋆a=5−5d

then,

\begin{gathered}S_{11} = \frac{11}{2} (2a + (11 - 1)d) \\ S_{11} = \frac{11}{2}(2(5 - 5d) + 10d) \\ S_{11} = \frac{11}{2} (10 - 10d + 10d) \\ S_{11} = \frac{11}{2} \times 10 = 11 \times 5 \\ S_{11} = 55\end{gathered}

S

11

=

2

11

(2a+(11−1)d)

S

11

=

2

11

(2(5−5d)+10d)

S

11

=

2

11

(10−10d+10d)

S

11

=

2

11

×10=11×5

S

11

=55

So, your answer is 55

Hope it helps you...