2 Marks Questions

Question 512 Marks

Find the sum of the following arithmetic progressions:
$41,36,31, \ldots .$. to 12 terms.

Answer

In an A.P. let first term $= a$, common difference $= d$, and there are n terms. Then, sum of n terms is,
$S_{n}=\frac{n}{2}\{2 a+(n-1) d\}$
Given expression is,
$41,36,31 \ldots . . . . \text { to } 12 \text { terms. }$
First term (a) = 41
Common difference $( d )=36-41=-5$
Sum of $n^{\text {th }}$ terms $s_n$, given $n=12$
$S_{12}=\frac{n}{2}(2 a(n-1) d)$
$=\frac{12}{6}(2.41+(12-1)-5)$
$=6(82+11(-5))$
$=6(27)$
$=162$
$\therefore S_{12}=162$

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Question 522 Marks

Find the $9^{\text {th }}$ term from the end of the A.P. $5,9,13, \ldots . . ., 185$.

Answer

Here,
First term, $a = 5$
Common difference, $d=9-5=4$
Last term, $I = 185$
$n ^{\text {th }}$ term from the end $=1-(n-1) d$
$9^{\text {th }}$ term from the end $=185-(9-1) 4=185-8 \times 4=185-32=153$

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Question 532 Marks

Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
$p, p + 90, p + 80, p + 270, .....$ where $p = (999)^{999}$

Answer

In the given problem, we are given various sequences.
We need to find out that the given sequences are an A.P. of not and then find its common difference (d),
$p, p + 90, p + 180, p + 270, .....$ where $p = (999)$
In the given sequence
$a_1 = p, a_2 = p + 90, a_3 = p + 180, a_4 = p + 270$
Check the condition
$a_2 - a_1 = a_3 - a_2$
$p + 90 - p = p + 180 - P - 90$
$90 = 180 - 90$
$90 = 90$

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Question 542 Marks

For what value of p are 2p + 1, 13, 5p - 3 are three consecutive terms of an A.P.?

Answer

$\because$ 2p + 1, 13, 5p - 3 are consecutive terms of an A.P.
$\therefore$ c.d. = 13 - 2p - 1, = 5p - 3 - 13
⇒ 5p + 2p = 13 - 1 + 13 + 3
$\Rightarrow\ 7\text{p}=28\Rightarrow\ \text{p}=\frac{28}{7}=4$
Hence p = 4

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Maths 2 Marks Questions