Question

The sum of 5th and 8th terms of an A.P is 37 and its 11th term is 32. Find the A.P

Answer 1

Solution:-

Formula:-

$\implies \boxed{ \rm \: A_n = a + (n - 1)d}$

:- 5th term

$\to \rm \: A_5 = a + (5 - 1)d \\ \to \: \rm A_5 \: = a + 4d \: \: \: \: \: \: \: \: \: \: \: \:$

:- 8th term

$\to \rm \: A_8 = a + (8- 1)d \\ \to \: \rm A_8\: = a + 7d \: \: \: \: \: \: \: \: \: \: \: \:$

:- 11th term

$\to \rm \: A_{11}= a + (11 - 1)d \\ \to \: \rm A_{11}\: = a + 10d \: \: \: \: \: \: \: \: \: \: \: \:$

Given:-

Sum of 5th and 8th term is 37

$\implies \: \rm \: a + 4d + a + 7d = 37$

$\rm \: \implies2a + 11d = 37 \: \: ..(i)eq$

and 11th term is 32

$\rm \implies \: a + 10d = 32 \: \: \: ...(ii)eq$

Now using substitution method

Taking (ii)eq

$\rm \: \implies \: a = 32 - 10d \: \: ..(iii)eq$

now substitute value of a on (i) eq

$\rm \: \implies \: 2a + 11d = 37$

$\rm \implies2(32 - 10d) + 11d = 37$

$\rm \: \implies \: 64 - 20d + 11d \: = 37$

$\rm \implies \: - 9d \: = 37 - 64$

$\rm \implies \: - 9d = - 27$

$\rm \implies \: d = 3$

Now put the value of d on (iii)eq

$\rm \: \implies \: a = 32 - 10d \: \:$

$\rm \: \implies \: a = 32 - 10 \times 3$

$\rm \implies \: a = 32 - 30$

$\rm \implies \: a = 2$

Ap is

:- 2 , 5 , 8 , 11 .........

Answer 2

$\huge\bf\underline\pink{Solution}$

Given that

Sum of 5th term and 8th term is 37

5th term = a+4d

8th term = a +7d

Given that Sum of 5th term and 8th term is 37

So,

a + 4d + a + 7d = 37

2a + 11d = 37 ______ eq 1

And also given thar 11term of Ap = 32

So, a + 10d = 32___eq 2

a = 32-10d Sub in eq 1

2(32-10d)+11d = 37

64-20d+11d = 37

64-9d =37

64 - 37 = 9d

27 = 9d

d = 3

Substuite value of d eq 2

a + 30 = 32

a =2

AP = 2,5,8,11....... Hence common difference is 3