Question

(ii) How many terms of the AP 16, 14, 12,... are needed to give the sum 60? Explain why do we get two answers. on the same​

Answer 1

Step-by-step explanation:

Given :-

The A.P. is 16,14,12,...

To find :-

The number of terms in the A.P. gives the sum is 60.

Solution :-

Given Arithmetic Progression is 16,14,12,...

First term (a) = 16

Common difference (d) = 14-16 = -2

Given sum (S) = 60

We know that

The sum of the first 'n' terms of an AP is

S = (n/2)[2a+(n-1)d]

On applying these values in the above formula then

=> 60 = (n/2)[2(16)+(n-1)(-2)]

=> 60 = (n/2)[32+(-2n+2)]

=> 60 = (n/2)(32-2n+2)

=> 60 = (n/2)(34-2n)

=> 60 = (n/2)(2×(17-n))

=> 60 = (n)(17-n)

=> 60 = 17n-n²

=> 60-17n+n² = 0

=> n²-17n+60 = 0

=> n²-5n-12n+60 = 0

=> n(n-5)-12(n-5) = 0

=> (n-5)(n-12) = 0

=> n-5 = 0 (or) n-12 = 0

=> n = 5 (or) n = 12

The value of n is 5 or 12.

Answer :-

The number of terms is 5 or 12 gives the sum 60 in the A.P.

Check:-

Case-1:-

If n = 5 then Sum of 5 terms of the AP

=> (5/2)[2(16)+(5-1)(-2)]

=> (5/2)[32+(4×-2)]

=> (5/2)(32-8)

=> (5/2)(24)

=> (5×24)/2

=> 5×12

=> 60

Case -2:-

If n = 12 then Sum of 12 terms of the AP

=> (12/2)[2(16)+(12-1)(-2)]

=> (6)[32+(11×-2)]

=> (6)(32-22)

=> (6)(10)

=> 60

Verified the given relations in the given problem.

Used formulae:-

→ The sum of the first 'n' terms of an AP is S = (n/2)[2a+(n-1)d]

• n = number of terms
• a = first term
• d = common difference
• S = sum of the first n terms of an A.P.

Answer 2

$\underline{\huge \tt{ \red{A} \blue{n} {S} \green{w} \pink{E} \orange{r}}}$

$\\ \bigstar\boxed{ \bf{ s_{n} = \frac{n}{2} \{2a + (n -1 )d \} }}$

AP is 16,14,12,...

Here ,

• a = 16
• d = 14-16 = -2
• S(n) = 60

$\longmapsto\sf \: 60 = \dfrac{n}{2} \{ 2a + (n - 1) (- 2)\} \\ \\ \longmapsto\sf \: 60 \times 2 = n \{2(16) - 2n + 2 \} \\ \\ \longmapsto\sf \: 120 = 32n - 2 {n}^{2} + 2n \\ \\ \longmapsto\sf 120 = 34n - 2 {n}^{2} \\ \\ \longmapsto\sf \: 2 {n}^{2} - 34n + 120 = 0$

Dividing both sides by 2 , we get..

$\longmapsto\sf \: {n}^{2} - 17n + 60 = 0 \\ \longmapsto\sf \: {n}^{2} - 12n - 5n + 60 = 0 \\ \longmapsto\sf \: n(n - 12) - 5(n - 12) = 0 \\ \longmapsto\sf (n - 5)(n - 12) = 0 \\ \boxed{\sf \red{n = 5}} \: \: \: \: \: \: \boxed{ \sf \red{n = 12}}$

We get two answers because as we continue the AP , there are negative terms.