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

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

• Fourth term of an A.P is 10. Eleventh term is one more than 3 times the 4th term.

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}$

• Sum of all 25 terms in the A.P.

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}$

First,

• Fourth term of A.P is 10.

This implies that,

• $\sf a_4$ = 10

• a + 3d = 10

Next,

• 11th term is one more than 3 times the 4th term.

This implies that,

• $\sf a_{11} \: = \: 3a_4 \: + \: 1$

Now,

We know that,

• $\sf a_4$ = 10

Substituting the values,

• $\sf a_{11} \: = \: 3 \: * \: 10 \: + \: 1$

• $\sf a_{11} \: = \: 30 \: + \: 1$

• $\sf a_{11} \: = \: 31$

• a + 10d = 31

• a = 31 - 10d

Now,

Substituting a = 31 - 10d in a + 3d = 10,

• a + 3d = 10

• 31 - 10d + 3d = 10

• - 10d + 3d = 10 - 31

• - 7d = - 21

• $\sf d \: = \: \dfrac{- \: 21}{- \: 7}$

• $\sf d \: = \: \dfrac{\cancel{-} \: \cancel{21}}{\cancel{-} \: \cancel{7}}$

• d = 3

Now,

Substituting d = 3 in a + 3d = 10,

• a + 3d = 10

• a + 3 * 3 = 10

• a + 9 = 10

• a = 1

Now,

• There are 2 ways of finding the sum of the first 25 terms.

1st Method:-

• $\sf a_{25} \: = \: ?$

• a + 24d

• 1 + 24 * 3

• 1 + 72

• 73

We know that,

$\boxed{\pink{\sf S_n \: = \: \dfrac{n}{2} \Big(a \: + \: l \Big)}}$

Here,

• n = 25

• a = 1

• l = 73

Substituting the values,

• $\sf S_{25} \: = \: \dfrac{25}{2} \Big(1 \: + \: 73 \Big)$

• $\sf S_{25} \: = \: \dfrac{25}{2} (74)$

• $\sf S_{25} \: = \: \dfrac{25}{2} \: * \: 74$

• $\sf S_{25} \: = \: \dfrac{25}{\cancel{2}} \: * \: \cancel{74}$

• $\sf S_{25} \: = \: 25 \: * \: 37$

• $\sf S_{25} \: = \: 925$

2nd Method,

We know that,

$\boxed{\pink{\sf S_{n} \: = \: \dfrac{n}{2} \Big[2a \: + \: (n \: - \: 1) \: d \Big]}}$

Here,

• n = 25

• a = 1

• d = 3

Substituting the values,

• $\sf S_{25} \: = \: \dfrac{n}{2} \Big[2a \: + \: (n \: - \: 1) \: d \Big]$

• $\sf S_{25} \: = \: \dfrac{25}{2} \Big[2 \: * \: 1 \: + \: (25 \: - \: 1) \: 3 \Big]$

• $\sf S_{25} \: = \: \dfrac{25}{2} \Big[2 \: + \: (24) \: 3 \Big]$

• $\sf S_{25} \: = \: \dfrac{25}{2} \Big[2 \: + \: (24) \: 3 \Big]$

• $\sf S_{25} \: = \: \dfrac{25}{2} \Big[2 \: + \: 72 \Big]$

• $\sf S_{25} \: = \: \dfrac{25}{2} \Big[74 \Big]$

• $\sf S_{25} \: = \: \dfrac{25}{2} \: * \: 74$

• $\sf S_{25} \: = \: \dfrac{25}{\cancel{2}} \: * \: \cancel{74}$
• $\sf S_{25} \: = \: 25 \: * \: 37$

• $\sf S_{25} \: = \: 925$

Since, the answer is same by both the methods.

Therefore,

$\bullet{\leadsto} \: \underline{\boxed{\blue{\texttt{Sum of the first 25 terms of the A.P = 925.}}}}$

$\Large{\bf{\purple{\mathfrak{\dag{\underline{\underline{Extra \: Information:-}}}}}}}$

Arithmetic progression:-

• It is a list of numbers in which each term is obtained by adding a fixed number d to the preceding term, expect to the 1st term.

Formulas:-

$\sf a_n \: = \: a \: + \: (n \: - \: 1) \: d$

• where,

• $\underline{\sf a_n}$ is the nth term.

• a is the first term of the A.P.

• n is the number of a term/terms.

• d is the common difference.

$\sf S_n \: = \: \dfrac{n}{2} \Big(a \: + \: l \Big)$

• where,

• $\underline{\sf S_n}$ is the sum of n terms.

• n is the number of terms.

• a is the 1st term of the A.P.

• l is the last term of the A.P (based on n).

$\sf S_{n} \: = \: \dfrac{n}{2} \Big[2a \: + \: (n \: - \: 1) \: d \Big]$

• where,

• $\underline{\sf S_n}$ is the sum of n terms.

• n is the number of terms.

• a is the 1st term of the A.P.

• d is the common difference.

Answer 2

Answer:

Given :-

• The fourth term of AP is 10.
• 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 AP
• a = First term of AP
• d = Common difference
• n = Number of terms

Solution :-

$\mapsto$ The fourth term of AP 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,

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

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

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

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

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

Again, 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 AP (a) = 1

According to the question by using the formula we get,

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

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

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

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

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

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

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

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

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