Question

If 1st and 6th terms of an AP are –12 and 8 and, sum of n terms is 120, then the number of terms is:​

Answer 1

$\huge\underline{\overline{\mid\star{\sf{\pink{Solution :—}}\star\mid}}}$

Given ,

• First term (a) → -12
• Sixth term (a6) → 8
• Sum of n terms (Sn) → 120

To find : —

Number of terms (n)

Here,

a = -12

And a6 = 8

=> a + 5d = 8

=> -12 + 5d = 8

=> 5d = 20

=> d = 4

Now,

Sn = 120

=> n/2 × {a + a+(n-1)d} = 120

=> n/2 × { -12 -12 + (n-1)4 } = 120

=> n{-24 + 4n -4} = 240

=> n{ -28 + 4n} = 240

=> -28n + 4n² = 240

=> 4n² - 28n -240 = 0

=> n² - 7n - 60 = 0

=> n² -12n + 5n - 60 = 0

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

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

$\implies \: n \: + 5 = 0 \: \: \: \: or \:\: \: n - 12 = 0 \\ \implies \: n \: = - 5 \: \: \: \: \: \: \: \: \: \implies \: n \: = 12 \\ ( unacceptable)$

$\Large{\colorbox{green}{\underline{\underline{♣Answer♣:—\:\:Number\:of\: terms\:= 12} }}}</p><p>$

Answer 2

Answer:

n=12

Step-by-step explanation:

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

120=n/2[2(-12x2)+(n-1)(4)]

120=n/2[-24+4n-4]

120=n/2[-28+4n]

240=n[-28+4n]

240=4n^2-28n

0=4n^2-28n-240

0=4[n^2-7n-60]

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

n=12