Question

in an ap if the 5th and 12th term of an ap whose second and third term are 14 and 18 respectively​

Answer 1

$\sf{\red{\underline{\huge{\red{Correct\:Question}}}}}$

In an A.P. find the 5th and 12th term whose 2nd and 3rd term are 14 and 18 respectively.

Given:-

• 2nd term = 14
• 3rd term = 18

To find:-

5th and 12th term.

Solution:-

2nd term = 14

We know,

$\sf{a_n = a + (n-1)d}$

= $\sf{a_2 = a + (2-1)d}$

= $\sf{14 = a + d}$

=> $\sf{a+d = 14\longrightarrow[i]}$

Now,

3rd term = 18

= $\sf{a_3 = a + (3-1)d}$

= $\sf{18 = a + 2d}$

= $\sf{a+2d = 18\longrightarrow[ii]}$

Subtracting Eq.[i] from [ii]

$\sf{(a+2d)-(a+d) = 18-14}$

= $\sf{a+2d-a-d = 4}$

= $\sf{d = 4}$

Putting the value of d in eq.[i]

$\sf{a+d = 14}$

= $\sf{a + 4 = 14}$

=> $\sf{a = 14-4}$

=> $\sf{a = 10}$

Therefore,

First term (a) = 10

Common difference (d) = 4

To find 5th term and 12th term:-

For 5th term:-

$\sf{a_n = a + (n-1)d}$

= $\sf{a_5 = 10 + (5-1)\times4}$

= $\sf{a_5 = 10 + 4\times4}$

= $\sf{a_5 = 10+16}$

= $\sf{a_5 = 26}$

Now,

For 12th term,

$\sf{a_{12} = 10+(12-1)\times4}$

= $\sf{a_{12} = 10 + 11\times4}$

= $\sf{a_{12} = 10 + 44}$

= $\sf{a_{12} = 54}$

Therefore,

$\sf{\underline{\boxed{\sf{5th\:term = 26}}}}$

$\sf{\underline{\boxed{\sf{12th\:term = 54}}}}$

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General Information:-

• To find nth term:- $\sf{a_n = a+(n-1)d}$

• To find the sum of nth term = $\sf{\dfrac{n}{2}[2a+(n-1)d]}$

• Where a = first term and d = common difference.

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