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Standard XII

Mathematics

nth Term of A.P

Let a1,a2,....

Question

Let $a_{1}, a_{2}, . . . a_{3 n}$ be an arithmetic progression with $a_{1} = 3$ and $a_{2} = 7$ . If $a_{1} + a_{2} + . . . + a_{3 n} = 1830$ then what is the smallest positive integer $m$ such that $m (a_{1} + a_{2} + . . . + a_{n}) > 1830 ?$

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Solution

$a_{1} = 3$ $∴ d = a_{2} - a_{1} = 7 - 3 = 4$
$a_{2} = 7$
$a_{1} + a_{2} + a_{3} + . . . . + a_{3 n} = 1830$
$\frac{3 n}{2} [2 \times 3 + (3 n - 1) \times 4] = 1830$
$n (6 + 12 n - 4)$ =
$12 n^{2} + 2 n = 1220$
$6 n^{2} + n - 610 = 0$
$(6 n + 61) (n - 10) = 0$
$∴ n = 10$
As n cannot be negative
$∴ m (3 + 7 + 11 + . . . . + 39) > 1830$
$m \times \frac{10}{2} (2 \times 3 + (10 - 1) \times 4) > 1830$
$m \times 5 (6 + 36) > 1830$
$m \times 5 \times 42 > 1830$
$m > 8.714$
$∴ m = 9$

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Similar questions

Let $a_{1}, a_{2} \dots a_{3 n}$ be an arithmetic progression with $a_{1} = 3$ and $a_{2} = 7$ . If $a_{1} + a_{2} + \dots + a_{3 n} = 1830$ , then what is the smallest positive integer m such that $m (a_{1} + a_{2} + \dots + a_{n}) > 1830$ ?

Q. Let

a_{1}, a_{2}, a_{3}, \dots, a_{100}

be an arithmetic progression with

a_{1} = 3

and

S_{p}

is sum of 100 terms . For any integer

n

with

1 \leq n \leq 20

, let

m = 5 n

. If

\frac{S_{m}}{S_{n}}

does not depend on

n

, then

a_{2}

If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ are in arithmetic progression, where $a_{1} > 0$ for all i.
Prove that $\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \dots + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} = \frac{n - 1}{\sqrt{a_{1}} + \sqrt{a_{n}}}$

Q. Let

a_{1}, a_{2}, a_{3}, \dots \dots, a_{100}

be an arithmetic progression with

a_{1} = 3

and

S_{p} = \sum_{i = 1}^{p} a_{i}, 1 \leq p \leq 100

. For any integer n with

1 \leq n \leq 20

, let m = 5n. If

\frac{S_{m}}{S_{n}}

does not depend on n, then

a_{2}

is equal to___

The n^th term (general term) of an arithmetic progression, a₁, a₂, a₃, …, a_n, … where a₂ –a₁ = a₃ – a₂ = …. = d is:

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Let a1,a2,...a3n be an arithmetic progression with a1=3 and a2=7. If a1+a2+...+a3n=1830 then what is the smallest positive integer m such that m(a1+a2+...+an)>1830?

a1=3 ∴d=a2−a1=7−3=4a2=7a1+a2+a3+....+a3n=18303n2[2×3+(3n−1)×4]=1830n(6+12n−4)=12n2+2n=12206n2+n−610=0(6n+61)(n−10)=0∴n=10As n cannot be negative∴m(3+7+11+....+39)>1830m×102(2×3+(10−1)×4)>1830m×5(6+36)>1830m×5×42>1830m>8.714∴m=9

Let $a_{1}, a_{2}, . . . a_{3 n}$ be an arithmetic progression with $a_{1} = 3$ and $a_{2} = 7$ . If $a_{1} + a_{2} + . . . + a_{3 n} = 1830$ then what is the smallest positive integer $m$ such that $m (a_{1} + a_{2} + . . . + a_{n}) > 1830 ?$