Question

If the 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term, find the A.P​

Answer 1

Step-by-step explanation:

Given:-

• The 5th term of an A.P. is 31.

• 25th term is 140 more than the 5th term.

To Find:-

• The A.P

Solution:-

Case 1:-

5th term of an A.P. is 31

$\sf \implies a_{5} = 31$

$\sf \implies a + 4d= 31.....(i)$

Case 2:-

25th term is 140 more than the 5th term.

$\sf \implies a_{25} = 140 + a_{5}$

$\sf \implies a + 24d = 140 + (a + 4d)$

$\sf \implies a + 24d = 140 + 31$

$\sf \implies a + 24d = 171.....(ii)$

Equation (ii) - (i):-

$\sf \implies a + 24d - (a + 4d) = 171 - 31$

$\sf \implies a + 24d - a - 4d = 140$

$\sf \implies 20d = 140$

$\sf \implies d = \dfrac{140}{20}$

$\sf \implies d = 7$

Substitute d = 7 in equation (i):-

$\sf \implies a + 4d = 31$

$\sf \implies a + 4(7) = 31$

$\sf \implies a = 31 - 28$

$\sf \implies a = 3$

The first term = a = 3

The second term = a + d = 3 + 7 = 10

The third term = a + 2d = 3 + 14 = 17

Therefore:-

$\underline{\boxed{\therefore\textsf{\textbf{The \: AP = 3, 10, 17, 24.....}}}}$

Answer 2

Answer:

$\huge{ \underline{ \rm{ \large{ \pink{Given:}}}}}$

• 5th term of AP = 31
• ${ \bold{ \sf{ a_{5} = 31 }}}$
• 25th term if AP = 140 more than the 5th term
• ${ \bold{ \sf{ a_{25} = 140 + a_{5} }}}$

$\huge{ \underline{ \large{ \red{ \rm{}Find:}}}}$

• AP??

$\huge{ \underline{ \large{ \rm{ \green{Solution:}}}}}$

${ \sf{ a_{5} = 31}}$

From the formula ⤵

$\: \: \: \: \: \: \: \: \: \: { \boxed{ \sf{ a_{n} = a + (n - 1)d }}}$

⇒ a + (5 - 1) d = 31

⇒ a + 4d = 31

⇒ a = 31 - 4d ...... (1)

${ \sf{a_{25} = 140 + a_{5}}}$

⇒ a + (25 - 1) d = 140 + 31

⇒ a + 24d = 171......(2)

Now substitute the value of equation (1) in equation (2):

⇒ 31 - 4d + 24d = 171

⇒ 31 + 20d = 171

⇒ 20d = 171 - 31

⇒ 20d = 140

⇒ d = 7

So, the common difference is 7

From equation (1),

⇒ a = 31 - 4d

⇒ a = 31 - 4×7

⇒ a = 31 - 28

⇒ a = 3

Lets find the AP:-

First term = a = 3

Second term = a + d = 3 + 7 = 10

Third term = a + 2d = 3 + 2×7 = 3+14 = 17

Fourth term = a + 3d = 3 + 3×7 = 3+21 = 24

Thus the required AP is 3 , 10, 17, 24....