Question

the fourth term of an Ap is 10. If eleventh term is one more three times of the fourth term find the sum of its 25 terms​

Answer 1

Fourth term of an AP, a₄ = 10

11th term = 1 + 3 × Fourth term

To Find: Sum of its 25 terms.

Given that,

• a₄ = 10 or a + 3d = 10 ...(i)

Also,

⇒ a₁₁ = 1 + 3×a₄

⇒ a + 10d = 1 + 3(a + 3d)

⇒ a + 10d = 1 + 3a + 9d

⇒ 2a - d = -1 ...(ii)

Multiply eq.(ii) by 3, we get

⇒ 6a - 3d = -3 ...(iii)

Adding (i) & (iii), we get

⇒ a + 3d + 6a - 3d = 10 + (-3)

⇒ 7a = 7

⇒ a = 1

Substitute [a = 1] in eq.(i) ,

⇒ 1 + 3d = 10

⇒ 3d = 9

⇒ d = 3

Now, we have

• n = 25
• a = 1
• d = 3

The sum of first n terms of an AP is given by,

⇒ Sₙ = n/2 [2a + (n - 1)d ]

⇒ S₂₅ = 25/2 { 2(1) + 24×3 }

⇒ S₂₅ = 25/2 ( 2 + 72 )

⇒ S₂₅ = 25/2 × 74

⇒ S₂₅ = 25 × 37

⇒ S₂₅ = 925

Hence, the sum of the first 25 terms of the given AP is 925.

Answer 2

Answer:

Given :-

• The fourth term of an AP is 10.
• If eleventh term is one more three times of the fourth term.

To Find :-

• What is the sum of its 25 terms.

Formula Used :-

$\longmapsto \sf\boxed{\bold{\pink{S_n =\: \dfrac{1}{2}\bigg[2a + (n - 1)d\bigg]}}}$

where,

• $\sf S_n$ = Sum of a term of A.P
• a = First term of A.P
• d = Common difference
• n = Number of terms

Solution :-

$\mapsto$ The fourth term of A.P is 10.

$\implies \sf a_4 =\: 10$

$\implies \sf\bold{\purple{a + 3d =\: 10\: ------\: (Equation\: No\: 1)}}$

$\mapsto$ Eleventh term is one more three times of the fourth term.

$\implies \sf a_{11} =\: 1 + 3 \times a_4$

Here we have,

• $\sf a_4$ = a + 3d

$\implies \sf a + 10d =\: 1 + 3 \times a + 3d$

$\implies \sf a + 10d =\: 1 + 3(a + 3d)$

$\implies \sf a + 10d =\: 1 + 3a + 9d$

$\implies \sf - 1 =\: 3a - a + 9d - 10d$

$\implies \sf\bold{\purple{- 1 =\: 2a - d\: ------\: (Equation\: No\: 2)}}$

Now, by multiplying the equation no 1 by 3 we get,

$\implies \sf - 1 =\: 2a - d$

$\implies \sf\bold{\purple{- 3 =\: 6a - 3d\: ------\: (Equation\: No\: 3)}}$

Now, by adding the equation no 1 and 3 we get,

$\implies \sf a \cancel{+ 3d} + 6a \cancel{- 3d} =\: 10 + (- 3)$

$\implies \sf a + 6a =\: 10 - 3$

$\implies \sf 7a =\: 7$

$\implies \sf a =\: \dfrac{\cancel{7}}{\cancel{7}}$

$\implies \sf\bold{\green{a =\: 1}}$

Now, by putting the value of a = 1 in the equation no 2 we get,

$\implies \sf - 1 =\: 2a - d$

$\implies \sf - 1 =\: 2(1) - d$

$\implies \sf - 1 =\: 2 - d$

$\implies \sf - 1 - 2 =\: - d$

$\implies \sf \cancel{-} 3 =\: \cancel{-} d$

$\implies \sf 3 =\: d$

$\implies \sf\bold{\green{d =\: 3}}$

Now, we have to find the sum of its 25 terms :

Given :

• Number of terms (n) = 25
• Common difference (d) = 3
• First term of A.P (a) = 1

According to the question by using the formula we get,

$\implies \sf S_{25} =\: \dfrac{25}{2}\bigg[2(1) + (25 - 1)3\bigg]$

$\implies \sf S_{25} =\: \dfrac{25}{2}\bigg[2 + 24 \times 3\bigg]$

$\implies \sf S_{25} =\: \dfrac{25}{2}\bigg[2 + 72\bigg]$

$\implies \sf S_{25} =\: \dfrac{25}{2}\bigg[74\bigg]$

$\implies \sf S_{25} =\: \dfrac{25}{\cancel{2}} \times {\cancel{74}}$

$\implies \sf S_{25} =\: 25 \times 37$

$\implies\sf\bold{\red{S_{25} =\: 925}}$

$\therefore$ The sum of its 25 terms is 925.

$\rule{300}{2}$

Extra Formula :

$\leadsto \sf\bold{\pink{a_n =\: a + (n - 1)d}}$

where,

• $\sf a_n$ = nth term of the sequence
• a = First term of the sequence
• d = Common difference