Question

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Find the first term and common difference of an AP whose 14th term is 40 and the sum of first fourteen terms is 287​

Answer 1

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Answer 2

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

An A.P. whose 14th term is 40 and the sum of first fourteen terms is 287.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The first term and the common difference.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We know that formula of an A.P.;

$\boxed{\bf{a_{n}=a+(n-1)d]}}}$

• a is the first term
• d is the common difference.
• n is the term of an A.P.

A/q

$\longrightarrow\sf{a_{14}=40}\\\\\longrightarrow\sf{a+(14-1)d=40}\\\\\longrightarrow\sf{a+13d=40............................(1)}$

&

We know that formula of the sum of an A.P;

$\boxed{\bf{S_{n}=\frac{n}{2}\big[2a+(n-1)d\big]}}}$

$\longrightarrow\sf{S_{14}=287}\\\\\\\longrightarrow\sf{\cancel{\dfrac{14}{2} }\bigg[2a+(14-1)d\bigg]=287}\\\\\\\longrightarrow\sf{7\big[2a+13d\big]=287}\\\\\\\longrightarrow\sf{2a+13d=\cancel{\dfrac{287}{7} }}\\\\\\\longrightarrow\sf{2a+13d=41.......................(2)}$

Subtracting equation (1) and equation (2),we get;

$\longrightarrow\sf{a\cancel{+13d}-2a \cancel{-13d}=40-41}\\\\\longrightarrow\sf{a-2a=-1}\\\\\longrightarrow\sf{\cancel{-}a=\cancel{-}1}\\\\\longrightarrow\sf{\orange{a=1}}$

Putting the value of a in equation (1),we get;

$\longrightarrow\sf{1+13d=40}\\\\\longrightarrow\sf{13d=40-1}\\\\\longrightarrow\sf{13d=39}\\\\\longrightarrow\sf{d=\cancel{\dfrac{39}{13} }}\\\\\longrightarrow\sf{\orange{d=3}}$

Thus;

The first term of an A.P. (a) = 1 and common difference (d) = 3.