Question

The 5th term of an ap is 17/6 and the 9th term is 25/6. What is the 12th term? Select one:

a. 33/6

b. 30/6

c. 31/6

d. 21/6

Answer 1

Answer:

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Answer 2

The 5th term of an AP is 17/6. The 12th term is 31/6

Solution:

Given that fifth term of an A.P $= a_{5} = \frac{17}{6}$

And 9th term of an A.P $= a_{9} = \frac{25}{6}$

Using formula of nth term,

$\mathrm{a}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$

where "a" = first term of A.P

"d" is the common difference between the terms

Using the above formula for 5th term,

$a_{5}=a+(5-1) d=a+4 d$

Given that 5th term is equal to $\frac{17}{6}$

$a + 4d = \frac{17}{6}$

=> 6a + 24d = 17     ------ (1)

Similarly for 9th term,

$\begin{array}{l}{a_{9}=a+(9-1) d=a+8 d} \\\\ {=>a+8 d=\frac{25}{6}}\end{array}$

=> 6a + 48d = 25     --------(2)

On subtracting equation (2) from (1) we get

(6a +48d) – (6a + 24d)  = 25 -17

=> 24d = 8

=> d = $\frac{2}{6}$

Using (1),

6a + 24 x $\frac{2}{6}$ = 17

=>6a = 17 – 8

=> a = $\frac{9}{6}$

For 12th term, we get,

$\mathrm{a}_{12}=\mathrm{a}+(12-1) \times \mathrm{d}=\frac{9}{6}+11 \mathrm{x} \frac{2}{6} = \frac{31}{6}$

Hence 12th term of A.P is $\frac{31}{6}$ Hence option C is correct