Find the value of x such that $25 + 22 + 19 + 16 + . . . + x = 112$

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Solution

The correct option is B $7$
Series is $25 + 22 + 19 + 16 + . . . . .$
which is in A.P. where,
First term $a = 25$
and Common difference, $d = a_{2} - a_{1} = 22 - 25 = - 3$

Sum of n terms, $S_{n} = 112$
where $n$ th term is $x$

$S_{n} = \frac{n}{2} {2 a + (n - 1) d}$

So,

$\frac{n}{2} {2 (25) + (n - 1) (- 3)} = 112$

$n {50 - 3 n + 3} = 224$

$3 n^{2} - 53 n + 224 = 0$
$3 n^{2} - 21 n - 32 n + 224 = 0$
$(n - 7) (3 n - 32) = 0$

$n = 7$ or $n = \frac{32}{7}$

$n$ is no. of terms which cannot be a fraction. So, $n = 7$

$x = a_{7} = a + 6 d$
$x = 25 + 6 (- 3) = 7$

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Solve:
(i)25 + 22 + 19 + 16 + ... + x = 115
(ii) 1 + 4 + 7 + 10 + ... + x = 590

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(ii)

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(iv)

\frac{13}{6} = \frac{- 65}{x}

(v)

\frac{16}{x} = - 4

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\frac{- 48}{x} = 2

Using the method of completion of squares find one of the roots of the equation $2 x^{2} - 7 x + 3 = 0$ . Also, find the equation obtained after completion of the square.

Q. Solve :

25 + 22 + 19 + 16 + . . . . + x = 115

Q. If

f (x - 1) = 3 x^{2} + 5 x + 6

, find the value of f(x+1).

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Find the value of x such that 25+22+19+16+...+x=112

Find the value of x such that $25 + 22 + 19 + 16 + . . . + x = 112$