Question

How many terms of the A.P. 16, 14, 12, are needed to give the sum 60? Explain .... why do we get two answers.​

Answer 1

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

Given AP series is 16,14, 12, -----

It means

First term of an AP, a = 16

Common difference of an AP, d = 14 - 16 = - 2

Further given that,

Sum of the series, Sₙ = 60.

Let assume that number of terms in an AP series be n

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic progression is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the progression.

n is the no. of terms.

d is the common difference.

So, on substituting the values, we get

$\rm \: 60 = \dfrac{n}{2} \bigg(2 \times 16 + (n - 1)( - 2)\bigg)$

$\rm \: 60 = \dfrac{n}{2} \bigg(32 - 2n + 2\bigg)$

$\rm \: 60 = \dfrac{n}{2} \bigg(34 - 2n\bigg)$

$\rm \: 60 = n(17 - n)$

$\rm \: 60 = 17n - {n}^{2}$

$\rm \: {n}^{2} - 17n + 60 = 0$

$\rm \: {n}^{2} - 12n - 5n + 60 = 0$

$\rm \: n(n - 12) - 5(n - 12) = 0$

$\rm \: (n - 12)(n - 5) = 0$

$\rm\implies \:n = 12 \: \: or \: \: n = 5$

Explanation of double answer

As given series is 16, 14, 12, ... is a decreasing AP series.

and can be written further as

16, 14, 12, 10, 8, 6, 4, 2, 0, - 2, -4, - 6

So, it implies 9th term of an AP is 0 and onwards this, 10th, 11th and 12th term are negative.

These negative terms when added to 6th, 7th and 8th term, they cancel out each other.

So, number of terms needed be 5 or 12.

$\rule{190pt}{2pt}$

Additional Information :-

↝ 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.

Answer 2

Step-by-step explanation:

The answer is 5 or 12.

This is your answer.