Question

In an arithmetic sequence 4th term is 31and 8th term is59.find 25th term

Answer 1

Step-by-step explanation:

a+3d=31

a+7d=59

By solving the above given equations we get d=7

&substituting d=7 in a+3d=31 we get a=10

25 th term = a+24d =10+168 =178

Answer 2

Answer:

$\large{ \bull \: \: \green{\underline{ \underline{ \red{ \boxed{ \color{pink}{ \mathfrak{25th \: term \: = 178}}}}}}}}$

Step-by-step explanation:

Given :

$\\ \sf\pink{where}{\begin{cases} \sf 4th \: term \: is \: 31 \\ \\ \sf 8th \: term \: is \: 59\end{cases}}$

To Find :

• the 25th term

Solution :

$\underline{ \bf{As \: we \: know : }}$

$\large{ \star{ \underline{ \boxed{ \color{red}{ \mathbf{t_n = a + (n - 1)d}}}}}}$

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We can also write the terms in different way.

$\\ \bf \: \to t_4 = a + 3d$

$\\ \to \bf 31 = a + 3d \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \small{( {eq}^{1}) }$

And,

$\\ \to \bf \: t _8 = a + 7d$

$\\ \bf \to59 = a + 7d \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \small{( {eq}^{2})}$

Subtracting eq¹ from eq²

$\\ \tt59 = {a} + 7d \\ \tt31 = a + 3d \\ - - - - - - - \\ \tt28 = 4d \\ \tt d = \cancel\frac{28}{4} \\ \pink{\star} \boxed{ \underline{ \tt \: d = 7}}$

Putting the value of d on eq¹

$\\ \blue{ \dashrightarrow} \tt \: 31 = a + 3(7)$

$\\ \blue{ \dashrightarrow} \tt31 = a + 21$

$\\ \blue{ \dashrightarrow} \tt \: a = 31 - 21$

$\\ \blue{ \dashrightarrow} \pink{ \star}{ \boxed{ \underline{\tt \: a = 10}}}$

Now we will use the formula of finding a term

$\\ \tt \: \longmapsto \: \color{magenta}{t_n = a + (n - 1)d}$

$\\ \longmapsto \tt t_{25} = 10 + (25 - 1)7$

$\\ \longmapsto \tt t_{25} =10 + (24)7$

$\\ \longmapsto \tt t_{25} =10 + 168$

$\\ \longmapsto \tt t_{25} =178$

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$\underline{ \text{So the value of 25th term is 178}}$