Question

29. The sum of how many terms of the AP 8. 15. 22. ... is 1490?

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Answer 1

★ Given :-

• AP = 8. 15. 22.

★ Have to Find :-

• How many terms of AP taken to give sum of 1490.

★ Solution :-

First we have to know :-

➤ What is Sequence ?

• Order of numbers or terms in which they are arranged written by an.

➤ What is A.P ?

• A sequence is said to be an A.P if the difference of two consecutive terms is always same and this difference is called Common difference denoted by ' d ' and the first term of sequence is denoted by 'a'

Now, lets do the Question

From the given A.P we know

• First term ↬ A1 = 8
• Common Difference ↬ A2 - A1 = 15 - 8 = 7

Given:-

$\sf{a = 8}$

$\sf{d = 7}$

$\sf{S_n = 1490}$

Formula of sum of terms :

${\boxed{\bf{S_n = \dfrac{n}{2}[2a + (n-1)d]}}}$

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

$\implies \sf{1490 = \dfrac{n}{2}[2a + (n-1)\times 7]}$

$\implies \sf{1490 \times 2 = n[2(8) + 7n - 7]}$

$\implies \sf{2980 = n[16 + 7n - 7]}$

$\implies \sf{2980 = n[16 - 7 + 7n]}$

$\implies \sf{2980 = n[9 + 7n]}$

$\implies \sf{2980 = 9n + 7n^2}$

By splitting middle term method :

= $\sf{7n^2 + 9n - 2980 = 0}$

$\sf{7n^2 + (149 - 140)n - 2980 }$

$\sf{7n^2 + 149n - 140n - 2980 }$

$\sf{7n^2 - 140n + 149n - 2980 }$

= $\sf{7n(n - 20) + 149 (n - 20) = 0}$

= $\sf{(7n + 149)(n - 20) = 0}$

So,

• $\sf{n - 20 = 0}$
• $\sf{n = 20}$

Or,

• $\sf{7n +149 = 0}$
• $\sf{7n = -149 }$

$\dashrightarrow\:\:\sf {n= \dfrac{-149}{7}}$

Negative value of n is negligible.

N = 20

$\dashrightarrow\:\: \underline{ \boxed{\sf Number \: of \: Terms = 20}}$

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Answer 2

Answer:

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