Question

The sum of the 5th and the 7th term of an ap is 52 and the 10th term is 46 find the ap ​

Answer 1

Answer:

Given, sum of the 5th and 7th term is 52.

a+4d+a+6d=52⇒2a+10d=52⟶(1)

And, the 10th term is 46.

a+9d=46

multiplying it by 2 we get,

2a+18d=92⟶(2)

Solving equations (1) and (2) we get,

8d=40⇒d=5

Substituting the value of d in (1) we get,

2a+(10×5)=52 implies that2a=52−50 implies that a=1

Therefore, the required A.P. is,

1,6,11,16,21,26,31

Answer 2

$\\ \large \dag \underbrace{ \textsf{ \textbf{concept : - }}}$

here were are supposed to use the concept of Arithmetic Progressions in which we have few terms in a given sequence having a common difference between each consecutive term.

☞ a is the first term in any AP

☞ d is the common difference between each consecutive term.

☞ $\sf a_{n}$ is the nth term

$\\ \large \dag \underbrace{ \textsf{ \textbf{Understanding the Question : - }}}$

here, were are provided with the sum of 5th and 7th terms of the A.P which is equivalent to 52. Also we are given the 10th term of this AP which is 46 and we are asked to find the AP.

So for that firstly we'll use the formula for finding the value of $\sf a_{n}$ in the sum of 5th and 7th terms, through which we'll get one linear equation in terms of a and d.

We also have another linear equation which is provided in the form of 10th term. By solving the equations we'll get the values of a and d through which we can determine the required A.P.

$\\ \large \dag \underbrace{ \textsf{ \textbf{Formula to be used :- : - }}}$

$\\ \quad\leadsto \underline{ \boxed{ \pink{\sf a_{n} = a + (n - 1)d}}} \bigstar$

Let's get started !!

• 5th term + 7th term = 52

• 5th term can be written as : a + 4d

• 7th term can be written as : a + 6d

let's write the given statement in terms of "a" and "d"

$\\ \quad \rightarrow \tt (a + 4d) + (a + 6d) = 52$

$\\ \quad \rightarrow \tt 2a + 10d = 52$

$\\ \quad \rightarrow \tt \blue{a + 5d = 26 - - - (1)}$

Now, the second statement is as follows :-

• 10th term of this AP is 46

• 10th term can be written as : a + 9d

$\\ \quad \rightarrow \tt\green{a + 9d = 46 - - - (2)}$

Now, let's solve the above equations by elemenation method. Subtract the 1st equation from the 2nd equation and elemenate "a" to find the value of "d"

$\\ \quad \dashrightarrow \tt \orange{(a + 9d) - (a + 5d) = 46 - 26} \\ \\ \quad \dashrightarrow \tt\red{ \cancel{a} + 9d - \cancel{a} - 5d = 20} \\ \\ \quad \dashrightarrow \tt \orange{4d = 20} \\ \\ \quad \dashrightarrow \tt \red{d = 5} \bigstar$

Now, substitute the value of d in any of the equation to calculate "a"

$\\ \quad \rightarrow \tt \pink{a + 5d = 26} \\ \\ \quad \rightarrow \tt \purple{a + 5(5) = 26} \\ \\ \quad \rightarrow \tt\pink{a + 25 = 26} \\ \\ \quad \rightarrow \tt\purple{a = 1} \bigstar$

Now, as we've calculated the values for a and d, we can determine the A.P by adding the common difference in each consecutive term of the AP

• $\sf a_{1} = 1$

• $\sf a_{2} = 1 + 5 = 6$

• $\sf a_{3} = 6 + 5 = 11$

• $\sf a_{4} = 11 + 5 = 16$

Hence, the required AP is :-

$\\ \bigstar \underline{ \boxed{ \orange{\bf AP = 1,6,11,16...}}} \bigstar$

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