Question

Is it possible that the product of any two term in the Ap 9,16,23... a term in it? Justify your answer

Answer 1

AP - 9, 16, 23, ...

Let's try by using an example from the AP itself.

Consider the first two terms 9 and 16.

9*16 = 144

Let's see if 144 is a term in the AP.

From the AP,

• The first term (a) = 9.
• The common difference (d) = 16 - 9 = 7.

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

144 = a + (n - 1)d

144 = 9 + (n - 1)(7)

144 = 9 + 7n - 7

144 = 2 + 7n

142 = 7n

142/7 = n

Since the value of n isn't a definite one, it isn't a term of the AP.

Thus, the product of any two terms in the AP 9, 16, 23, ... isn't a term in it.

Answer 2

Answer:

$no \: it \: is \: not \: possible$

Step-by-step explanation:

Given----> An ap ,

9, 16, 23, ...........

To find ---->

$to \: check \: that \: it \: is \: possible \: that \: the \\ product \: of \: any \: two \: term \: is \: a \: term \: of \: ap \:$

Concept used----> 1)

$nth \: term \: of \: ap \: is \: \\ = a \: + (n - 1)d$

2)

$value \: of \: number \: of \: term \: can \: not \: be \: fractional$

Solution----> 1) please see the attachement

2)

$product \: of \: mth \: term \: and \: nth \: term = pth \: term$

$= > (7m + 2) \: (7n + 2) \: = a + (p - 1)d$

$= > 49mn + 14m + 14n + 4 = 9 + (p - 1)7$

$= > 49mn + 14(m + n) + 4 = 9 + 7p - 7$

$= > 7p + 2 = 49mn + 14(m + n) + 4$

$= > 7p = 49mn + 14(m + n) + 4 - 2$

$= > p = \dfrac{49mn + 14(m + n)}{7} + \dfrac{ {2} }{7}$

$= > p = \dfrac{7 (7mn + 2m + 2n)}{7} + \dfrac{2}{7}$

$= > p = 7mn + 2m + 2n + \dfrac{2}{7}$

$mn \: and \: (m + n) \: is \: integer \: but \: \dfrac{2}{7} is \: not \: integer \\ so \: p \: is \: not \: integer \: \\ so \: it \: is \: not \: possible \: product \: of \: any \: two \: term \: is \: itself \: a \: term \: of \: given \:ap$