Question

A sequence exercise. Find a1 and d ig given S3 and a10?
S3= -10.5
a10= 2.9​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:a\:=\:-\:4.3\:}\:,\:\red{\sf\:d\:=\:0.8\:}}}$

Step-by-step-explanation:

We have given that,

For an AP,

$\displaystyle{\bullet\:\sf\:S_{3}\:=\:-\:10.5}$

$\displaystyle{\bullet\:\sf\:a_{10}\:=\:2.9}$

We have to find a₁ i. e. a and d.

We know that,

$\displaystyle{\boxed{\pink{\sf\:S_n\:=\:\dfrac{n}{2}\:[\:2a\:+\:(\:n\:-\:1\:)\:d\:]\:}}}$

$\displaystyle{\implies\sf\:S_3\:=\:\dfrac{3}{2}\:[\:2a\:+\:(\:3\:-\:1\:)\:d\:]}$

$\displaystyle{\implies\sf\:-\:10.5\:=\:\dfrac{3}{2}\:[\:2a\:+\:2d\:]}$

$\displaystyle{\implies\sf\:\dfrac{-\:\cancel{10.5}\:\times\:2}{\cancel{3}}\:=\:2a\:+\:2d}$

$\displaystyle{\implies\sf\:-\:3.5\:\times\:2\:=\:2a\:+\:2d}$

$\displaystyle{\implies\sf\:2\:(\:a\:+\:d\:)\:=\:-\:3.5\:\times\:2}$

$\displaystyle{\implies\sf\:a\:+\:d\:=\:\dfrac{-\:3.5\:\times\:\cancel{2}}{\cancel{2}}}$

$\displaystyle{\implies\sf\:a\:+\:d\:=\:-\:3.5}$

$\displaystyle{\implies\sf\:a\:=\:-\:3.5\:-\:d}$

$\displaystyle{\implies\:\boxed{\sf\:a\:=\:-\:d\:-\:3.5}\sf\:\qquad\cdots\:(\:1\:)}$

Now, we know that,

$\displaystyle{\boxed{\blue{\sf\:a_n\:=\:a\:+\:(\:n\:-\:1\:)\:d\:}}}$

$\displaystyle{\implies\sf\:a_{10}\:=\:a\:+\:(\:10\:-\:1\:)\:d}$

$\displaystyle{\implies\sf\:2.9\:=\:a\:+\:9d}$

$\displaystyle{\implies\sf\:(\:-\:d\:-\:3.5\:)\:+\:9d\:=\:2.9\:\qquad\:\cdots\:[\:From\:(\:1\:)\:]}$

$\displaystyle{\implies\sf\:-\:d\:-\:3.5\:+\:9d\:=\:2.9}$

$\displaystyle{\implies\sf\:9d\:-\:d\:=\:2.9\:+\:3.5}$

$\displaystyle{\implies\sf\:8d\:=\:6.4}$

$\displaystyle{\implies\sf\:d\:=\:\cancel{\dfrac{6.4}{8}}}$

$\displaystyle{\implies\:\boxed{\green{\sf\:d\:=\:0.8\:}}}$

By substituting this value in equation ( 1 ), we get,

$\displaystyle{\sf\:a\:=\:-\:d\:-\:3.5\:\qquad\cdots\:(\:1\:)}$

$\displaystyle{\implies\sf\:a\:=\:-\:(\:0.8\:)\:-\:3.5}$

$\displaystyle{\implies\sf\:a\:=\:-\:0.8\:-\:3.5}$

$\displaystyle{\implies\:\boxed{\purple{\sf\:a\:=\:-\:4.3\:}}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:a\:=\:-\:4.3\:}\:,\:\red{\sf\:d\:=\:0.8\:}}}}$