Question

$If \: T_{1} \: T_{2} \: T_{3} \: are \:in \: AP \: Such \; that \: T_{1} + T_{2} \:..... T_{25} = T_{26} + T_{27} +.....T_{40}$ and first term is three then value of common difference of AP is ?

Answer 1

GIVEN :-

• $\: T_{1} \: T_{2} \: T_{3} \: are \:in \: AP \: Such \; that \: T_{1} + T_{2} \:..... T_{25} = T_{26} + T_{27} +.....T_{40}$ and first term is three .

TO FIND :-

• value of common difference of AP .

SOLUTION :-

Let d = common difference .

Given that ,

$T_{1} + T_{2} \:..... T_{25} = T_{26} + T_{27} +.....T_{40}$

$\: S_{25} = T_{1} + T_{2} \:..... T_{25}\:...(T_{1} + T_{2} \:..... T_{25})\: +T_{26}+T_{27}+...T_{40}$

》 Sum of 25th term = Sum of 40th term - Sum of 25th term

》 2 × sum of 25th term = sum of 40th term

$\implies \: 2( \frac{25}{2} (2a + (n - 1)d)) = \frac{40}{2} (2a + (n - 1)d \\ \\ \\ \\ \implies \: 25(2 \times 3 + 24d) = 20(6 + 39d) \\ \\ \\ \implies \: 25(6 + 24d) = 20(6 + 39d) \\ \\ \\ \implies \: 150 + 600d = 120 + 780d \\ \\ \\ \implies \: 150 - 120 = 780d - 600d \\ \\ \\ \implies \: 30 = 180d \\ \\ \\ \implies \: d = 180 \div 30 = \frac{1}{6}$

Hence the answer is ,

$\rightarrow \: \boxed{d = \frac{1}{6} }$

HOPE IT HELPS :)

Answer 2

EXPLANATION.

⇒ T₁, T₂, T₃, . . . . . are in ap.

⇒ T₁ + T₂ + T₃ + . . . . . + T₂₅ = T₂₆ + T₂₇ + T₂₈ + . . . . . + T₄₀.

First term : a = 3.

As we know that,

Sum of nth terms of an ap.

⇒ Sₙ = n/2[2a + (n - 1)d].

Using this formula in this question, we get.

We can write expression as,

Sum of first 25 terms = Sum of first 40 terms - Sum of first 25 terms.

⇒ S₂₅ = S₄₀ - S₂₅.

⇒ 2 x S₂₅ = S₄₀.

⇒ 2 x [25/2[2a + (25 - 1)d] = 40/2[2a + (40 - 1)d].

⇒ 25[2a + 24d] = 20[2a + 39d].

⇒ 5[2a + 24d] = 4[2a + 39d].

Put the value of a = 3 in the equation, we get.

⇒ 5[2(3) + 24d] = 4[2(3) + 39d].

⇒ 5[6 + 24d] = 4[6 + 39d].

⇒ 30 + 120d = 24 + 156d.

⇒ 30 - 24 = 156d - 120d.

⇒ 6 = 36d.

⇒ d = 1/6.

∴ value of common difference : d = 1/6.

                                                                                                               

MORE INFORMATION.

Supposition of terms in an A.P.

Three terms as : a - d, a, a + d.

Four terms as : a - 3d, a - d, a + d, a + 3d.

Five terms as : a - 2d, a - d, a, a + d, a + 2d.

Arithmetic progression (A.P) : if a is the first term and d is the common difference then A.P. can written as,

a + (a + d) + (a + 2d) + . . . . .