Question

Let a₁, a₂,.....,a₃₀ be an A.P., S = ³⁰∑ᵢ₌₁ aᵢ and T = ¹⁵∑ᵢ₌₁ a₍₂ᵢ₋₁₎ . If a₅ = 27a and S - 2T = 75, then a₁₀ is equal to (A) 52 (B) 57
(C) 47 (D) 42
[JEE Main 2019]

Answer 1

Option A: The value of $a_{10}$ is 52

Explanation:

Given that $a_1,a_2,....,a_{30$ be an A.P

Also, given that $S=^{30}\sum_{i=1}a_i$ and $T=^{15}\sum_{i=1}a_{(2i-1)}$

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

We need to determine the value of the term $a_{10}$

The value of the term $a_{10}$ can be determined using the formula,

$a_{10}=a+(n-1)d$

Since, $S=^{30}\sum_{i=1}a_i$

$S=a_1+a_2+....+a_{30}$

  $=\frac{30}{2} [2a+29d]$

$S=15[2a+29d]$

Also, $T=^{15}\sum_{i=1}a_{(2i-1)}$

$T=a_1+a_3+a_5+...+a_{29}$

   $=\frac{15}{2}(2a+28d)$

$T=15(a+14d)$

The value of $a_5$ is 27

$a_5=a+(5-1)d$

$27=a+4d$ -------------(1)

The value of $S-2T$ is 75

$15(2a+29d)-2[15(a+14d)]=75$

Simplifying, the terms, we get,

$30a+435d-30a-420d=75$

Adding the terms, we have,

$15d=75$

  $d=5$

Substituting $d=5$ in equation (1), we get,

$27=a+4(5)$

$27=a+20$

$7=a$

Now, we shall substitute the value of a and d in the formula,

$a_{10}=a+(n-1)d$

     $=7+(10-1)(5)$

     $=7+(9)(5)$

$a_{10}=52$

Thus, the value of $a_{10}$ is 52

Therefore, Option A is the correct answer.

Learn more:

(1) Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12

brainly.in/question/3820092

(2) Find "am" for an arithmetic progression if sum of "m" terms is given by sm = 2m2 + 5m

brainly.in/question/3081300

Answer 2

Option A: The value of a_{10}a

10

is 52

Explanation:

Given that a_1,a_2,....,a_{30 be an A.P

Also, given that S=^{30}\sum_{i=1}a_iS=

30

∑

i=1

a

i

and T=^{15}\sum_{i=1}a_{(2i-1)}T=

15

∑

i=1

a

(2i−1)

If a_5=27a

5

=27 and S-2T=75S−2T=75

We need to determine the value of the term a_{10}a

10

The value of the term a_{10}a

10

can be determined using the formula,

a_{10}=a+(n-1)da

10

=a+(n−1)d

Since, S=^{30}\sum_{i=1}a_iS=

30

∑

i=1

a

i

S=a_1+a_2+....+a_{30}S=a

1

+a

2

+....+a

30

=\frac{30}{2} [2a+29d]=

2

30

[2a+29d]

S=15[2a+29d]S=15[2a+29d]

Also, T=^{15}\sum_{i=1}a_{(2i-1)}T=

15

∑

i=1

a

(2i−1)

T=a_1+a_3+a_5+...+a_{29}T=a

1

+a

3

+a

5

+...+a

29

=\frac{15}{2}(2a+28d)=

2

15

(2a+28d)

T=15(a+14d)T=15(a+14d)

The value of a_5a

5

is 27

a_5=a+(5-1)da

5

=a+(5−1)d

27=a+4d27=a+4d -------------(1)

The value of S-2TS−2T is 75

15(2a+29d)-2[15(a+14d)]=7515(2a+29d)−2[15(a+14d)]=75

Simplifying, the terms, we get,

30a+435d-30a-420d=7530a+435d−30a−420d=75

Adding the terms, we have,

15d=7515d=75

d=5d=5

Substituting d=5d=5 in equation (1), we get,

27=a+4(5)27=a+4(5)

27=a+2027=a+20

7=a7=a

Now, we shall substitute the value of a and d in the formula,

a_{10}=a+(n-1)da

10

=a+(n−1)d

=7+(10-1)(5)=7+(10−1)(5)

=7+(9)(5)=7+(9)(5)

a_{10}=52a

10

=52

Thus, the value of a_{10}a

10

is 52

Therefore, Option A is the correct answer.