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Let $a_1, a_2,... ....., a_{30}$ be an AP, $S = \sum_{i = 1}^{30}a_i$ and $T = \sum_{i = 1}^{15}a_{(2i-1)}$ .

If $a_5 = 27$ and $S-2T = 75$

then $a_{10}$ is equal to:
Option 1)
52
Option 2)
57
Option 3)
47
Option 4)
42

Answers (1)

A admin

Sum of n terms of an AP -

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

and

Sum of n terms of an AP

$S_{n}= \frac{n}{2}\left [ a+l\right ]$

- wherein

$a\rightarrow$ first term

$d\rightarrow$ common difference

$n\rightarrow$ number of terms

Last term of an A.P.(l) -

$l= a+(n-1)d$

- wherein

If a series has n terms last term is $l$

Summation of terms of an AP

$S=a_{1}+a_{2}+a_{3}\cdots \cdots \cdots \cdots a_{30}$

$=\frac{30}{2}\left [ a_{1} +a_{30}\right ]=15\left ( a_{1}+a_{30} \right )$

or, $S=15\left ( a_{1}+a_{1} +29d\right )$

Now,

$T=a_{1}+a_{3}+a_{5}+\cdots \cdots \cdots \cdots a_{29}$

$=a_{1}+\left ( a+2d \right )+\left ( a+4d \right )+\cdots +\left ( a+14d \right )$

$=15a_{1}+2d\left ( 1+2+\cdots \cdots +14 \right )$

$T=15a_{1}+210d$

Now use,

$S-2T=75$

$\Rightarrow 15\left ( 2a_{1}+29d \right )-2\left ( 15a_{1}+210d \right )=75$

$\Rightarrow d=5$

Give $a_{5}=27=a_{1}+4d\Rightarrow a_{1}=7$

Now, $a_{10}=a_{1}+9d$

$=7+9\times 5=52$

Option 1)
52
Option 2)

57

Option 3)

47

Option 4)

42

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