Question

in an a.p 8 term is 16 and 16 term is 48 find the 25 term o​

Answer 1

Answer:

T8 = 16

a+7d = 16

T16 = 48

a + 15d = 48

a = 48 - 15d

48 - 15d + 7d = 16

-8d = 16-48

-8d = -32

d = 4

a = 48 - 15(4)

= 48 - 60

= -12

T25 = a + 24d

= (-12) + 24(4)

= -12 + 96

= 84

Answer 2

$\large\underline{\sf{Solution-}}$

Let assume that

↝ First term of an AP is a

and

↝ Common difference of an AP is d.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Tʜᴜs, According to given,

$\red{\rm :\longmapsto\:a_8 = 16 \: }$

$\rm :\longmapsto\:a + (8 - 1)d = 16$

$\rm \implies\:\boxed{ \tt{ \: a + 7d = 16 \: }} - - - - - (1)$

Also, Given that

$\red{\rm :\longmapsto\:a_{16} = 48 \: }$

$\rm :\longmapsto\:a + (16 - 1)d = 48$

$\rm \implies\:\boxed{ \tt{ \: a + 15d = 48 \: }} - - - - (2)$

On Subtracting equation (1) from equation (2), we get

$\rm :\longmapsto\:8d = 32$

$\rm \implies\:\boxed{ \tt{ \: d \: = \: 4 \: }} - - - - (3)$

Substitute equation (3) in equation (1), we get

$\rm :\longmapsto\:a + 7 \times 4 = 16$

$\rm :\longmapsto\:a + 28= 16$

$\rm :\longmapsto\:a= 16 - 28$

$\rm \implies\:\boxed{ \tt{ \: a \: = \: - \: 12 \: }} - - - - (4)$

Now, Consider

$\red{\rm :\longmapsto\:a_{25} \: }$

$\rm \: = \: \:a + (25 - 1)d$

$\rm \: = \: \:a + 24d$

$\rm \: = \: \: - 12 + 24 \times 4$

$\rm \: = \: \: - 12 + 96$

$\rm \: = \: 84$

$\red{\rm \implies\:\boxed{ \tt{ \: a_{25} \: = \: 84 \: }}}$

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Additional Information :-

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.