Question

The 5th and 11th terms of an arithmetic
progression are 18 and 24 respectively.
Find the value of the term lying exactly
in between these terms.​

Answer 1

Solution :-

Let :-

First term = a

Common difference = d

We know :-

$\bf a_n = a+(n-1)d$

$\bullet \sf \ 5^{th} \ term = 18$

 $\longrightarrow \sf a_5 = 18 \\\\\longrightarrow a+(5-1)d = 18 \\\\\longrightarrow a+4d = 18 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .....................(1)$

$\bullet \sf \ 11^{th} \ term = 24$

  $\longrightarrow\sf a_{11} = 24\\\\\longrightarrow \sf a+(11-1)d = 24 \\\\\longrightarrow a+10d = 24 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .....................(2)$

Eq (2) - Eq (1) :-

  $\longrightarrow \sf 6d = 6 \\\\\longrightarrow \bf d = 1$

Substitute the value of d in eq (1) :-

 $\longrightarrow \sf a+4 \times 1 = 18 \\\\\longrightarrow a+4= 18 \\\\\longrightarrow \bf a = 14$

Term lying exactly between 5th term and 11th term = 8th term

$\longrightarrow \sf a_8 = 14+( 8-1)\times 1 \\\\\longrightarrow a_8 = 14+7 \\\\\longrightarrow a_8 = 21$

$\bf 8^{th} \ term = 21$

Answer 2

Answer:

here , n=5 An= 18

a+(n-1)d=An

a+(5-1)d=18

a+(4)d=18

a+4d=18 ---------eq 1

Now, n=11, An=24

a+(n-1)d=24

a+(11-1)d=24

a+(10)d=24

a+10d=24 -------------eq 2

subtract eq 1 from eq 2:

a+10d=24

a+4d =18

_. _. _

_____________

0 + 6d= 6

6d=6

d= 6/6

d = 1

now, put the value of d in eq 1

a+4d=18

a+4(1)=18

a+4 =18

a =18-4

a = 16

Step-by-step explanation:

SINCE,WE FIND THE VALUE OF a=16 and d=1

NOW YOU CAN FIND THE VALUE OF THE TERM LYING EXACTLY IN BETWEEN THESE TERMS

HOPE IT WILL HELP YOU ☺️