Question

How manHow many terms of the arithmetic sequence 5,7,9,...,must be added to get 480?

Answer 1

Step-by-step explanation:

Here,

First term, a = 5

Common difference, d = 2

Sum of n terms = 140

= n/2 × [2a + (n-1)d]

140 = n/2 × [2(5) + (n-1)2]

140 = n/2 × (10 + 2n - 2)

140 = n/2 × (8 + 2n)

n² + 4n - 140 = 0

On factorisation,

(n-10) (n+14) are the factors

So,

n = 10 or n = -14

n = -14 is REJECTED. (no. of terms can't be negative)

Thus, x = 10 is Correct

Hence, 10 terms must be added of AP 5,7,9,... to get 140

Let me know if you have doubts especially regarding Physics, Chemistry or Maths.

Answer 2

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

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$\large{\green{\underline \bold{\tt{Given :-}}}}$

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• An AP 5,7,9,...,

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$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

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• Terms of the AP 5,7,9,..., must be added to get 480

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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$\underline{\bold{\texttt{We know that ,}}}$

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➠ $\sf S_n = \dfrac { n } { 2 } (2a + (n - 1)d)$ ------ (1)

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Where,

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• $\sf S_n$ = Sum of n terms

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• a = Common difference

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• n = Number of terms

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• d = Common difference

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$\underline{\bold{\texttt{For the given AP 5,7,9,...,}}}$

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• $\sf S_n$ = Sum of n terms = 480

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• a = First term = 5

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• n = Number of terms = n

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• d = Common difference = 7 - 5 = 2

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⟮ Putting these value in (1) ⟯

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➠ $\sf S_n = \dfrac { n } { 2 } (2a + (n - 1)d)$

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➜ $\sf 480 = \dfrac { n } { 2 } (2(5) + (n - 1)2)$

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➜ $\sf 480 = \dfrac { n } { 2 } (10 + 2n - 2)$

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➜ $\sf 480 = \dfrac { n } { 2 } (8 + 2n)$

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➜ $\sf 480 = n \times (4 + n)$

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➜ $\sf 480 = 4n + n ^2$

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➜ n² + 4n - 480 = 0

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⟮ Splitting the middle term ⟯

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➜ n² + 24n - 20n - 480 = 0

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➜ n(n + 24) -20(n + 24) = 0

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➜ (n + 24)(n - 20) = 0

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• n = -24
• n = 20

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As the number of term can't be negative hence n = 20

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∴ 20 term of the given AP must be added to get 480 as the sum

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