Question

How many terms of an ap 9 17 25 dotdotdot dotdotdot must be taken to give a sum of 636?

Answer 1

Answer:

636 can't be the sum of first n terms of the AP 9, 17, 25,...

Step-by-step explanation:

⇒ Finding sum of first n term,

$a = 9 \\ \\ d = 17 - 9 = 8 \\ \\ S_n = \frac{n}{2} [2a + (n - 1)d] \\ \\ S_n = \frac{n}{2} [2 \times 9 + (n - 1)8] \\ \\ S_n = \frac{n}{2}[18+8n-8] \\ \\ S_n=\frac{n}{2}[8n+10] \\ \\ S_n=4n^2+5n$

⇒ Taking S_n = 636,

$S_n=636 \\ \\ 4n^2+5n=636 \\ \\ 4n^2+5n-636=0 \\ \\ 4n^2+48n-53n-636=0 \\ \\ 4n(n+12)-53(n+12)=0 \\ \\ (n+12)(4n-53)=0 \\ \\ n=-12\ \ \ ; \ \ \ n=\frac{53}{4}$

⇒ Here we can conclude with the fact that 636 can't be the sum of first n terms of the AP 9, 17, 25,..., because both the values of n got here are not positive integers.

Answer 2

Answer:

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