JEE Main 2019 — Sequences And Series Question with Solution

Let the common difference = d

S = $$\sum _{i=1}^{30}{a}_i$$

= $$a$$₁ + $$a$$₂ + . . . . . + $$a$$₃₀

$$\therefore$$  S = $$\frac{30}{2}\left[a_1+a_{30}\right]$$

= 15 [$$a$$₁ + $$a$$₁ + 29d]

= 15 (2$$a$$₁ + 29d)

T = $$\sum _{i=1}^{15}{a}_{\left(2i-1\right)}$$

= $$a$$₁ + $$a$$₃ + . . . . . . + $$a$$₂₉

= $$\frac{15}{2}\left[a{}_1+a_{29}\right]$$

= $$\frac{15}{2}\left[a{}_1+a_1+28d\right]$$

= $$\frac{15}{2}\left[2a{}_1+28d\right]$$

= 15 ($$a$$₁ + 14d)

Given,

S $$-$$ 2T = 75

$$\Rightarrow$$  15(2$$a$$₁ + 29d) $$-$$ 2 $$\times$$ 15 ($$a$$₁ + 14d) = 75

$$\Rightarrow$$  30$$a$$₁ + 15 $$\times$$ 29d $$-$$ 30 $$a$$₁ $$-$$ 420d = 75

$$\Rightarrow$$  435d $$-$$ 420d = 75

$$\Rightarrow$$  15d = 75

$$\Rightarrow$$  d = 5

Given that,

$$a$$₅ = 27

$$\Rightarrow$$  $$a$$₁ + 4d = 27

$$\Rightarrow$$  $$a$$₁ + 20 = 27

$$\Rightarrow$$  $$a$$₁ = 7

$$\therefore$$  $$a$$₁₀ = $$a$$₁ + 9d

= 7 + 45

= 52