Question

The 6th term of an ap is 18 and 9th term is 12 find 15 th term

Answer 1

Let the 1st term be a and common difference be d

A.T.Q

$6th \: \: term \: = 18 \\ a + (6 - 1)d = 18 \\ a + 5d = 18 \: \: \: \: \: \: \: \: \: - (1) \\ \\ 9th \: term = 12 \\ a + (9 - 1)d = 12 \\ a + 8d = 12 \: \: \: \: \: \: - (2) \\ \\ subtracting \: (1) \: from \: (2) \: we \: have \\ \\ (a + 8d) - (a + 5d) = 12 - 18 \\ a + 8d - a - 5d = - 6 \\ 3d = - 6 \\ d = \frac{ - 6}{3} = - 2 \\ \\ substituting \: the \: value \: of \: d \: in \: equation \: (1) \: we \: have \\ \\ a + 5( - 2) = 18 \\ a - 10 = 18 \\ a = 28 \\ \\ now \: according \: to \: question \: \\ \\ 15th \: term \: = a + (15 - 1)d = a + 14d \\ \\ = 28 + 14( - 2) = 28 - 28 = 0 \\ \\ hence \: the \: 15th \: term \: of \: this \: ap \: is \: 0$

Answer 2

The 15th term of an AP is 0 if the 6th term of an AP is 18 and 9th term is 12.

Concept used:

• First with the help of given two terms of AP, we will form two equations in a and d and then solve them to find the value of a and d.

• By using these values of a and d in the general term of AP and find the term.

Formula used:

General or nth term of an  AP : $a_{n}$ = a + ( n - 1)d

Here, a = first term, d = common difference, n = number of terms

Given :

6th term of an AP, $a_{6}$ = 18, 9th term of an AP, $a_{9}$ = 12

To find :

Find the 15th term.

Solution:

Step 1 : Find equation for 6th term of  an AP by using the Formula of General or nth term of an  AP:

General or nth term of an  AP : $a_{n}$ = a + ( n - 1)d

For 6th term we have, $a_{6}$ = 18 and n = 6

$a_{6}$ = a + (6 - 1)d

$a_{6}$ = a + 5d

18 = a + 5d …....(1)

Step 2 : Find equation for 9th term of  an AP by using the Formula of General or nth term of an  AP:

General or nth term of an  AP : $a_{n}$ = a + ( n - 1)d

For 9th term we have , $a_{9}$ = 12 and n = 9

$a_{9}$ = a + (9 - 1)d

$a_{9}$ = a + 8d

12 = a + 8d …..(2)

Step 3: Subtracting eq. 1 from eq. 2

12 = a + 8d

18 = a + 5d

(-)   (-) (-)

----------------

-6 = 3d

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d = $\frac{-6}{3}$

d = -2

Step 4: Substitute the value of d in eq 1.

18 = a + 5d

18 = a + 5(-2)

18 = a - 10

18 + 10 = a

a = 28

Step 5 : Find 15th term by using the Formula of General or nth term of an  AP:

General or nth term of an  AP : $a_{n}$ = a + ( n - 1)d

We have a = 28 , d = -2 and n = 15

$a_{15}$ = a + (n - 1)d

$a_{15}$ = 28 + (15 - 1)(- 2)

$a_{15}$ = 28 + 14 × - 2

$a_{15}$ = 28 - 28

$a_{15}$ = 0

Hence, the 15 term of an AP is 0.

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