Question

Section D 12. An A.P consists-of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 327. Find the A.P. 13 The difference of two natural numbers is 5 and the difference of their reciprocals is 1/10. Find AP​

Answer 1

$\large\underline{\sf{Solution-12}}$

Given that, An A.P consists-of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 327.

So, n = 37, which is odd, it means three middle most terms are

$\rm \: a_{18} + a_{19} + a_{20} = 225$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic progression is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the progression.

n is the no. of terms.

d is the common difference.

Tʜᴜs,

$\rm \: a + 17d + a + 18d + a + 19d = 225$

$\rm \: 3a + 54d = 225$

$\rm \: 3(a + 18d) = 225$

$\rm \: a + 18d = 75 - - - (1)$

Also, given that Sum of last three terms is 327.

$\rm \: a_{37} + a_{36} + a_{35} = 327$

$\rm \: a + 36d + a + 35d + a + 34d = 327$

$\rm \: 3a + 105d = 327$

$\rm \: 3(a + 35d) = 327$

$\rm \: a + 35d = 109 - - - - (2)$

On Subtracting equation (1) from (2), we get

$\rm \: 17d = 34$

$\rm\implies \:d = 2$

On substituting d = 2 in equation (1), we get

$\rm \: a + 36 = 75$

$\rm \: a = 75 - 36$

$\rm\implies \:a = 39$

So, required AP series is 39, 41, 43, 45, 47, 49, ...

$\large\underline{\sf{Solution-13}}$

Let assume that

First natural number be x

Second natural number be x + 5

According to statement,

$\rm \: \dfrac{1}{x} - \dfrac{1}{x + 5} = \dfrac{1}{10}$

$\rm \: \dfrac{x + 5 - x}{x(x + 5)} = \dfrac{1}{10}$

$\rm \: \dfrac{5}{x(x + 5)} = \dfrac{1}{10}$

$\rm \: {x}^{2} + 5x = 50$

$\rm \: {x}^{2} + 5x - 50 = 0$

$\rm \: {x}^{2} + 10x - 5x - 50 = 0$

$\rm \: x(x + 10) - 5(x + 10) = 0$

$\rm \: (x + 10)(x - 5) = 0$

$\rm\implies \:x = 5 \: \: or \: \: x = - 10 \: \{rejected \}$

So, required natural numbers are 5 and 10.

Answer 2

Step-by-step explanation:

$\underline{ \underline{ \text{ \( \rm Sol_{n} \text{ : - 12}\) }}}$

Since, total number of terms

(n) =37 (Odd number)

$\text{\(\therefore \quad \) Middle term \( =\left(\dfrac{37+1}{2}\right) \) th term}$

So, the three middle most terms are 18 th, 19 th and 20 th.

By given condition,

Sum of the three middle most terms

$\[ \begin{aligned} &=& 225 \\ \\ & & \tt a_{18}+a_{19}+a_{20} & \tt=225 \\ \\ \Rightarrow & & \tt(a+17 d)+(a+18 d)+(a+19 d) & \tt=225& & \\ \\ \Rightarrow & & \tt 3 a+54 d &=225 \\ \\ \Rightarrow & & \tt a+18 d &=75 \end{aligned} \]$

and sum of the last three terms =429

$\[ \begin{aligned} \Rightarrow & & \tt a_{35}+a_{36}+a_{37} & \tt=429 \\ \\ \Rightarrow & & \tt(a+34 d)+(a+35 d) & \tt+(a+36 d) \\ \\ & & &=429 \\ \\ \Rightarrow & & \tt 3 a+105 d & \tt=429 \\ \\ \Rightarrow & & \tt a+35 d & \tt=143 \end{aligned} \]$

Solving (i) and (ii) , we have

$\\ \[ \tt a=3 \text { and } d=4 \]$

$\\ \\ \[ \begin{array}{l} \therefore \text { Required AP is } \tt a, a+d, a+2 d \text {, } \\ \tt a+3 d, \ldots \\ \tt \text { that is, } 3,3+4,3+2(4), 3+3(4), \ldots \\ \tt \text { or } 3,7,3+8,3+12, \ldots \text { or } 3,7,11, \\ \tt 15, \ldots \end{array} \]$