Question

How many terms of the Arithmetic Progression 45, 39, 33, ... must be

taken so that their sum is 180 ? Explain the double answer.​

Answer 1

$\huge\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Answer}}}}}}}} \\ \large\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Your~answer↓}}}}}}}}$

Given :-

• An AP series 45, 39, 33, ....... having sum 180

To find :-

• number of terms

Formula used

$\large\bold\green{Sum \: of \: n \: terms, S_n \: = \frac{n}{2}(2a + (n - 1) \times d) }$

where,

♡ a = first term of an AP

◇ d = Common Difference of an AP

◇ n = number of terms of an AP

Solution :-

Here,

First term of an AP, a = 45

Common Difference of an AP, d = 39 - 45 = - 6

Sum of terms of an AP = 180

Let number of terms of an AP be 'n'.

Using formula of sum of n terms,

$\small\bold\red{Sum \: of \: n \: terms, S_n \: = \frac{n}{2}(2a + (n - 1) \times d) }$

Substitute the values of a, d, n

$\small\bold\red{180 = \frac{n}{2} (2 \times 45 + (n - 1) \times ( - 6))} \\ \small\bold\red{180 \times 2= n(90 - 6n + 6)} \\ \small\bold\red{ = > 360 = n(96 - 6n)}$

$\small\bold\red{ = > 360 = 6n(16 - n)} \\ on \: dividing \: b y \: 6 \: on \: both \: sides\\ \small\bold\red{ = > \: 60 = 16n - {n}^{2} } \\ \small\bold\red{ = > {n}^{2} - 16n + 60 = 0} \\ \small\bold\red{}\small\bold\red{ = > {n}^{2} - 10n - 6n + 60 = 0 } \\ \small\bold\red{ = > \: n(n - 10) - 6(n - 10) = 0} \\ \small\bold\red{ = > \: (n - 10)(n - 6) = 0} \\ \small\bold\red{ = > \: \: \: \: \: \: n \: = 10 \: \: \: \: or \: \: \: \: \: n = 6\: \: \: }$

$\huge \fcolorbox{black}{cya}{♛Hope it helps U♛}$