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Standard X

Mathematics

General Form of an AP

Find the nth ...

Question

Find the nth term and 9th term of the sequence 3, 7, 11, 15.........

4 n - 1

and

T_{9} = 35

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4 n + 1

and

T_{9} = 37

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2 n - 1

and

T_{9} = 35

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4 n - 1

and

T_{9} = 39

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Solution

The correct option is C $4 n - 1$ and $T_{9} = 35$
We know,
$n^{t h}$ term $a_{n} = a + (n - 1) d$

$a =$ First term
$d =$ Common difference
$n =$ number of terms
$a_{n} = n^{t h}$ term

Here,
$a = 3 a n d d = 7 - 3 o r 11 - 7 = 4$

$a_{n} = a + (n - 1) d = 3 + (n - 1) 4$
$a_{n} = 4 n - 1$

For $9$ th term put $n = 9$
$T_{9} = a_{9} = 4 \times 9 - 1 = 35$

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t_{7} = \frac{1}{9}

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t_{9} = \frac{1}{7}

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If $t_{n}$ represents $n^{t h}$ term of an A. P., $t_{2} + t_{5} - t_{3} = 10$ and $t_{2} + t_{9} = 17$ , find its first term and its common difference.

Q. If

t_{n}

, represents

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t_{2} + t_{5} - t_{3} = 10

and

t_{2} + t_{9} = 17

, then find its first term and its common difference.

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then find the value of

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Find the nth term and 9th term of the sequence 3, 7, 11, 15.........

The correct option is C 4n−1 and T9=35We know,nth term an=a+(n−1)da= First termd= Common differencen= number of termsan=nth termHere,a=3 and d=7−3 or 11−7=4an=a+(n−1)d=3+(n−1)4an=4n−1For 9th term put n=9T9=a9=4×9−1=35