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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