Question

find the middle terms of an A. P 9,15,21,27,......183​

Answer 1

Question:

Find the middle terms of an A. P.

$9,15,21,27,......183$

To find:

★ To find the middle terms.

Answer:

$Middle \: terms \: are \: \underline{ \sf \pink{\bf 93 \: and \: 99}}.$

Given:

★ Consecutive terms of an A.P. is given,

$\sf9,15,21,27,......183$

Step-by-step explanation:

$9,15,21,27,......183$

$First \: term \: \underline{ (a) = 9}$

$Common \: difference \underline{(d) = t_{2} - t_{1}}$

$\star \: t _{2} = 15$

$\star \: t _{1} = 9$

$d = 15 - 9$

$\underline{ \: d = 6 \: }$

$Last \: term \: \: \underline{(l) = 183}$

$no.of \: terms \: \underline{ n = \frac{l - a}{d} + 1}$

Now substituting the values,

$\implies \frac{183 - 9}{6} + 1$

$\implies 29 + 1$

$\implies \boxed{ n = 30}$

$\therefore \: There \: are \: terms \: \underline{ \: \bf30 \: }.$

$Middle \: terms \: are \: \underline{ \bf t _{15} \: and \: t _{16}}.$

${n}^{th} \: term \: \boxed{ t_{n} = a + (n - 1)d}$

$\implies t _{15} = a + 14d$

$\implies t _{15} = 9 + 14(6)$

$\implies 9 + 84$

$\implies \boxed{ t _{15} = 93}$

$\longrightarrow t _{16} = a + 15d$

$\longrightarrow t _{16} = 9 + 15(6)$

$\longrightarrow 9 + 90$

$\longrightarrow \boxed{t _{16} = 99}$

$\therefore Middle \: terms \: are \: \underline{ \bf 93 \: and \: 99}.$

Formula used:

$\bigstar \: n = \frac{l - a}{d} + 1$

$\bigstar \: t_{n} = a + (n - 1)d$

More information:

Arithmetic Progression :-

In a sequence the difference between two consecutive terms are equal, it is called Arithmetic Progression .

General form :-

a , a + 2d , a + 3d , a+4d ,......

Answer 2

Answer:

don't know buddy sorry....