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General Form of an AP

Find the comm...

Question

Find the common difference and write the next four terms of the AP :
$- 1$ , $\frac{1}{4}$ , $\frac{3}{2}$ ,....

d =

\frac{5}{2}

;

a_{4}

\frac{11}{4}

a_{5}

\frac{16}{4}

a_{6}

\frac{21}{4}

a_{7}

\frac{26}{4}

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d =

\frac{5}{4}

;

a_{4}

\frac{11}{4}

a_{5}

\frac{16}{4}

a_{6}

\frac{21}{4}

a_{7}

\frac{26}{4}

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d =

\frac{5}{4}

;

a_{4}

\frac{11}{4}

a_{5}

\frac{16}{4}

a_{6}

\frac{21}{4}

a_{7}

\frac{25}{4}

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d =

\frac{5}{2}

;

a_{4}

\frac{11}{4}

a_{5}

\frac{16}{4}

a_{6}

\frac{21}{4}

a_{7}

\frac{25}{4}

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Solution

The correct option is B

$d = \frac{5}{4}$ ; $a_{4} =$ $\frac{11}{4}$ , $a_{5} =$ $\frac{16}{4}$ , $a_{6}$ = $\frac{21}{4}$ , $a_{7} =$ $\frac{26}{4}$

Here, $a = - 1$
Therefore, common difference,
$d = \frac{1}{4} - (- 1)$
$\Rightarrow d = \frac{1}{4} + 1$
$\Rightarrow d = \frac{1 + 4}{4}$
$\Rightarrow d = \frac{5}{4}$
$\Rightarrow d = \frac{5}{4}$ ---(1)
The $n^{th}$ term of the A.P is $a_{n} = a + (n - 1) d$
$\Rightarrow a_{4} = a + 3 d$
$= - 1 + (3 \times \frac{5}{4})$
$= - 1 + \frac{15}{4}$
$= \frac{- 4 + 15}{4}$
$= \frac{11}{4}$
$∴ a_{4} = \frac{11}{4}$ ---(2)
$a_{5} = a + 4 d$
$= - 1 + (4 \times \frac{5}{4})$
$= - 1 + 5$
$= 4$
$∴ a_{5} = 4 = \frac{16}{4}$ ---(3)
$a_{6} = a + 5 d$
$= - 1 + (5 \times \frac{5}{4})$
$= - 1 + \frac{25}{4}$
$= \frac{- 4 + 25}{4}$
$= \frac{21}{4}$
$∴ a_{6} = \frac{21}{4}$ ---(4)
$a_{7} = a + 6 d$
$= - 1 + (6 \times \frac{5}{4})$
$= - 1 + \frac{30}{4}$
$= \frac{- 4 + 30}{4}$
$= \frac{26}{4}$
$= \frac{13}{2}$
$∴ a_{7} = \frac{26}{4} = \frac{13}{2}$ .---(5)

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Similar questions

Suppose $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ are integers such that $\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}$ where 0 $\leq$ a< j for j= 2,4,5,6,7.

The sum $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$ is

Q. Suppose

a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}

are integers such that

\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}

where

0 \leq a < j

for

j = 2, 4, 5, 6, 7.

The sum

a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}

Q. Suppose

a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}

are intagers such that

\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}

where

0 \leq a_{j} < j

for

j = 2, 3, 4, 5, 6, 7

. The sum

a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}

is-

The sum $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$ is

Q. If

\frac{3 x^{2} - 2 x + 4}{(x - 1)^{2}} = \frac{A_{1}}{x + 1} + \frac{A_{2}}{(x + 1)^{2}} + \frac{A_{3}}{(x + 1)^{3}} + \frac{A_{4}}{(x + 1)^{4}} + \frac{A_{5}}{(x + 1)^{5}} + \frac{A_{6}}{(x + 1)^{6}}

, then

(A_{1} + A_{3} + A_{5}, A_{2} + A_{4} + A_{6}) =

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Find the common difference and write the next four terms of the AP :−1, 14, 32,....

Find the common difference and write the next four terms of the AP :
$- 1$ , $\frac{1}{4}$ , $\frac{3}{2}$ ,....