Question

18. Given that sum of the 8th and 10th terms of an AP is 48 and the sum of the 8th and 12th terms is
44. Find the first three terms of the A.P?​

Answer 1

AP = 40 , 38 , 36 ...

$\bf\huge\underline{\underline{ Given :- }}$

• 8th term + 10th term == 48

• 8th term + 12th term == 44

$\bf\huge\underline{\underline{ To \: Find:- }}$

• First three terms of this AP .

$\bf\huge\underline{\underline{ Answer :- }}$

If a , b, c are in AP , $\sf\red{ b = \frac{a + c }{2}}$

• 9th term = $\frac{8th + 10th }{2}$

• $\red{\implies}$ 9th term = $\frac{48 }{2}$

• $\red{\implies}$ 9th term = $24$

$\rule{10cm}{0.05cm}$

• 10 th term = $\frac{8th + 12th }{ 2}$

• $\red{\implies}$ 10th term = $\frac{44 }{ 2}$

• $\red{\implies}$ 10th term = $22$

$\rule{10cm}{0.05cm}$

• 9th term = 24

• 10th term = 22

• common difference = 22 - 24 = -2

$\rule{10cm}{0.05cm}$

• 1 st term = 9th term - 8d

• $\red{\implies}$ 1st term = 24 + 16

• $\red{\implies}$ 1st term = 40

$\rule{10cm}{0.05cm}$

• 1st term = 40

• 2nd term = 38

• 3rd term = 36

$\rule{10cm}{0.05cm}$

Answer 2

Step-by-step explanation:

As we know that

$\tt \: a _{n} = a + (n - 1)d$

So,

$\tt \: a _8 = a + (8 - 1)d \\ \leadsto \tt \: a_8 \: = a + 7d - - - (i)$

And also,

$\tt \: a_{10 }= a + (10 - 1)d \\ \leadsto \tt \: a_{10} = a + 9d - - - (ii)$

Similarly

$\tt \: a_8 =a + (8 - 1)d \\ \tt \leadsto \: a_8 = a + 7b - - - (iii)$

Also

$\tt \: a_{12} = a + (12 - 1)d \ \\ \tt \leadsto a _{12} = a + 11d - - - (iv)$

As it is given that sum of 8th and 10th term is 48.

so,

$\tt \: a_8 + a_{10} = 48 \\ \tt \leadsto a + 7d + a + 9d = 48 \\ \tt \leadsto2a + 16d = 48 - - - (v)$

And also it is given that sun of 8th and 12th place is 44. So,

$\tt \: a_8 + a_{12} = 44 \\ \tt \leadsto \: a + 7d + a + 11d = 44 \\ \tt \leadsto \: 2a + 18d = 44 \\ \tt \leadsto \: a = \frac{44 - 18d}{2} - - - (vi)$

$\tt \: 2 \bigg( \frac{44 - 18d}{2} \bigg) + 16d = 48 \\ \tt \leadsto \: 44 - 18d + 16d = 48 \\ \tt \leadsto \: - 2d = 4 \\ \tt \leadsto \: d = - 2$

From vi

$\tt \: a = \frac{44 - (18 \times - 2)}{2} \\ \tt \: a = \frac{80}{2} \\ \tt \: a = 40$

1 st term=40

Second term =40+(-2)=38

Third term=38+(-2)=36