Question

If the fifth term of an arithmetic progression is 17 and the first term is 1, then it's 10th term is?​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the Concept of Arithmetic Progression has been used. We see that we are given the Firth term and First term of AP and we need to find the Tenth term of AP. By using 5th and 1st term, we can find the common difference. After finding the Common Difference, we can find the 10th term also.

Let's do it !!

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★ Formula Used :-

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

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★ Solution :-

Given,

» First Term = a = 1

» Fifth Term = a₅ = 17

~ For the Common Difference (d) ::

• Let the common difference be d.

We know that,

$\\\;\sf{\green{:\rightarrow\;\;a_{5}\;=\;\bf{a\;+\;4d}}}$

And,

$\\\;\sf{\green{:\rightarrow\;\;a_{5}\;=\;\bf{17}}}$

Now combining both the equations, we get

$\\\;\sf{\green{:\rightarrow\;\;a\;+\;4d\;=\;\bf{17}}}$

By applying the value of a,

$\\\;\sf{:\rightarrow\;\;1\;+\;4d\;=\;\bf{17}}$

$\\\;\sf{:\rightarrow\;\;4d\;=\;\bf{17\;-\;1}}$

$\\\;\sf{:\rightarrow\;\;4d\;=\;\bf{16}}$

$\\\;\sf{:\rightarrow\;\;4d\;=\;\bf{\dfrac{16}{4}}}$

$\\\;\bf{:\rightarrow\;\;Common\;Difference,\;d\;=\;\bf{\orange{4}}}$

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~ For the 10th term of the A.P. ::

We know that, nth term is given as

$\\\;\sf{:\rightarrow\;\;a_{n}\;=\;\bf{a\;+\;(n\;-\;1)d}}$

Now replacing n by 10, to get 10th term

$\\\;\sf{:\Longrightarrow\;\;a_{10}\;=\;\bf{1\;+\;(10\;-\;1)4}}$

$\\\;\sf{:\Longrightarrow\;\;a_{10}\;=\;\bf{1\;+\;(9)4}}$

$\\\;\sf{:\Longrightarrow\;\;a_{10}\;=\;\bf{1\;+\;36}}$

$\\\;\bf{:\Longrightarrow\;\;a_{10}\;=\;\bf{\blue{37}}}$

$\\\;\underline{\boxed{\tt{Hence,\;\;10^{th}\;\:term\;=\;\bf{\purple{37}}}}}$

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★ More to formulas to know :-

$\\\;\sf{\leadsto\;\;l\;=\;a\;+\;(n\;-\;1)(-d)}$

Here l shows the last term.

$\\\;\sf{\leadsto\;\;s_{n}\;=\;\dfrac{n}{2}\:(a\;+\;a_{n})}$

$\\\;\sf{\leadsto\;\;s_{n}\;=\;\dfrac{n}{2}\:(2a\;+\;(n\;-\;1)d)}$

$\\\;\sf{\leadsto\;\;s_{n}\;=\;\dfrac{n}{2}\:(a\;-\;l)}$

$\\\;\sf{\leadsto\;\;s_{n}\;=\;\dfrac{n}{2}\:(n\;+\;1)}$