Question

The sum of 5th and 9th terms of an ap is 30 if its 25th term is three times its 8th term find the AP.​

Answer 1

Step-by-step explanation:

Given:-

• The sum of 5th and 9th terms of an AP is 30.

• 25th term is three times its 8th term.

To Find:-

• The AP

Solution:-

Case 1:-

The sum of 5th and 9th terms of an AP is 30.

$\sf \implies a_{5} + a_{9} = 30$

$\sf \implies a + 4d + a + 8d = 30$

$\sf \implies 2a + 12d = 30......(i)$

Case 2:-

25th term is three times its 8th term.

$\sf \implies a_{25} = 3( a_{8})$

$\sf \implies a + 24d = 3(a + 7d)$

$\sf \implies a + 24d = 3a + 21d$

$\sf \implies a + 24d - 3a - 21d = 0$

$\sf \implies -2a + 3d = 0......(ii)$

Adding (i) and (ii):-

$\sf \implies 2a + 12d - 2a + 3d = 30$

$\sf \implies 15d = 30$

$\sf \implies d = \dfrac{30}{15}$

$\sf \implies d = 2$

Substitute d = 2 in equation (i)

$\sf \implies 2a + 12d = 30$

$\sf \implies 2a + 12(2) = 30$

$\sf \implies 2a = 30 - 24$

$\sf \implies 2a = 6$

$\sf \implies a = \dfrac{6}{2}$

$\sf \implies a = 3$

The first term = a = 3

The second term = a + d = 3 + 2 = 5

The third term = a + 2d = 3 + 4 = 7

$\large\underline{\boxed{\sf \therefore The \: AP = 3,5,7,9.....}}$

Answer 2

$\huge{\boxed{\rm{\red{Question}}}}$

The sum of 5th and 9th terms of an ap is 30 if its 25th term is three times its 8th term find the AP.

$\huge{\boxed{\rm{\red{Answer}}}}$

${\bigstar}\large{\boxed{\sf{\pink{Given \: that}}}}$

• The sum of 5th and 9th terms of an AP is 30.
• 25th term is three times its 8th term.

${\bigstar}\large{\boxed{\sf{\pink{To \: find}}}}$

• The AP.

${\bigstar}\large{\boxed{\sf{\pink{Solution}}}}$

$\large\purple{\texttt{In the first case}}$

$\large\orange{\texttt{The sum of 5th and 9th terms of an AP is 30}}$

↝ a⁵ + a⁹ = 30 {a⁵ means 5th term} ; {a⁹ means 9th term}

↝ a + 4d + a + 8d = 30

↝ 2a + 12d = 30 $\large\red{\texttt{Equation 1}}$

$\large\purple{\texttt{In the second case}}$

$\large\orange{\texttt{25th term is three times its 8th term}}$

↝ a²⁵ = 3(a⁸) {a²⁵ means 25th term}

↝ a + 24d = 3 (a+7d)

↝ a + 24d = 3a + 28d

↝ a + 24d - 3a - 21d = 0

↝ -2a + 3d = 0 $\large\red{\texttt{Equation 2}}$

Now we have to add $\large\red{\texttt{Equation 1}}$ and $\large\red{\texttt{Equation 2}}$

↝ 2a + 12d - 2a + 3d = 30

↝ 15d = 30

↝ d = 30/15

↝ d = 2

$\large{\boxed{\texttt{Now we have to substitute d = 2 in equation 1}}}$

↝ 2a + 12d = 30

↝ 2a + 12(2) = 30

↝ 2a + 24 = 30

↝ 2a = 30 - 24

↝ 2a = 6

↝ a = 6/2

↝ a = 3

Hence, the first term is 3

The second term is 3 + 2 = 5

The third term is 3 + 4 = 7

The AP is 3,5,7

Hope it's helpful.

Thank you :)