Question

4th term of ap is 17 and sum of 3rd& 7th is 42 find the first term and the common difference
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Answer 1

Step-by-step explanation:

Given:-

• The 4th term of an AP = 17

• The sum of 3rd and 7th terms is 42.

To Find:-

• The first term.

• The common difference.

Solution:-

Let " a " be the first term. And " d " be the common difference.

As, we know that :-

The nth term of an AP is:-

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

$\sf The \: 4th \: term = 17$

$\sf\longrightarrow a_{4} = 17$

$\sf \longrightarrow a + (4 - 1)d = 17$

$\sf \longrightarrow a + 3d = 17.....(i)$

$\sf The \: 3rd \: term + 7th \: term = 42$

$\sf \longrightarrow a_{3} + a_{7} = 42$

$\sf \longrightarrow a + 2d + a + 6d = 42$

$\sf \longrightarrow 2a + 8d = 42$

$\sf Dividing \: the\: whole \: equation\: by \: 2, \:we\: get[tex]</p><p></p><p>[tex]\sf \longrightarrow a + 4d = 21.....(ii)$

$\sf Equation \: (ii) - (i)$

$\sf \longrightarrow a + 4d - (a + 3d) = 21 - 17$

$\sf \longrightarrow a + 4d - a - 3d = 4$

$\longrightarrow\boxed{ \bf d = 4}$

$\sf Substitute \: d = 4\: in \: equation \: (i)$

$\sf \longrightarrow a + 3d = 17$

$\sf \longrightarrow a + 3(4) = 17$

$\sf \longrightarrow a = 17 - 12$

$\longrightarrow\boxed{ \bf a = 5}$

Therefore:-

$\underline{\boxed{ \bf The \: first \: term = 5}}$

$\underline{\boxed{ \bf Common \: difference = 4}}$