Question

the sum of 6th and 10th terms of an arithmetic sequence is 66 .
a) what is its 8th term
b)what is the sum of first 15 terms of this sequence ​

Answer 1

Answer:

$\bigstar{\bold{Eighth\:term(a_8)=33}}$

$\bigstar{\bold{Sum\:of\:15\:terms=495}}$

Step-by-step explanation:

$\Large{\underline{\rm{Given:}}}$

• Sum of 6th and 10th terms of an A.P is 66

$\Large{\underline{\rm{To\:Find:}}}$

• 8th term of the A.P
• Sum of first 15 terms of the A.P

$\Large{\underline{\rm{Solution:}}}$

➡ First we have to find the first term and common difference of the A.P

➡ The nth term of an A.P is given by,

   aₙ = a₁ + (n - 1) × d

    where a₁ is the first term

    d is the common difference

➡ Hence sixth term of the A.P is given by,

    a₆ = a₁ + 5d

➡ Also,

    a₁₀ = a₁ + 9d

➡ But by given,

    a₆ + a₁₀ = 66

➡ Hence,

    a₁ + 5d + a₁ + 9d = 66

    2a₁ + 14d = 66----(1)

➡ Dividing the whole equation by 2,

    a₁ + 7d = 33

➡ But we know that,

    a₁ + 7d = a₈

➡ Therefore,

   a₈ = 33

➡ Hence eighth term of the A.P is 33.

    $\boxed{\bold{Eighth\:term(a_8)=33}}$

➡ Now finding the sum of first 15 terms of the A.P

➡ The sum of n terms of an A.P is given by,

    $\tt S_n=\dfrac{n}{2} (2a_1+(n-1)\times d)$

➡ Substitute the datas,

    $\tt S_{15}=\dfrac{15}{2}(2a_1+14d)----(2)$

➡ Substitute equation 1 in equation 2,

    $\tt S_{15}=\dfrac{15}{2} \times 66$

    S₁₅ = 15 × 33

   S₁₅ = 495

➡ Hence the sum of 15 terms of the A.P is 495.

    $\boxed{\bold{Sum\:of\:15\:terms=495}}$