Question

how many terms are there in ap whose first term and 6th term are -12 and 8 respecttively and sum of all its term is 120

Answer 1

Given:-

• First term of an A.P = -12

• Sixth term of an A.P = 8

• Sum of all its term = 120

To Find:-

• No. of terms in an A.P

Formula Used:-

• ${\boxed{\bf{a_n=a+(n-1)d}}}$

• ${\boxed{\bf{S_n=\dfrac{n}{2}[2a+(n-1)d]}}}$

Here,

• $\bf S_n = 120$

• $\bf a_n =$ General term

• n = No. of terms

• a = First term

• d = Common Difference

Solution:-

As given,

$\sf :\implies\:a_6= 8$

$\sf :\implies\:a+(n-1)d= 8$

$\sf :\implies\:-12+(6-1)d= 8$

$\sf :\implies\:-12+5d= 8$

$\sf :\implies\:5d= 20$

$\sf :\implies\:d= \dfrac{20}{5}$

$\bf :\implies\:d= 4$

Now,

$\sf :\implies\:S_n=\dfrac{n}{2}[2a+(n-1)d]$

$\sf :\implies\:120=\dfrac{n}{2}[2\times(-12)+(n-1)4]$

$\sf :\implies\:120=\dfrac{n}{2}[-24+4n-4]$

$\sf :\implies\:120\times2=n[4n-28]$

$\sf :\implies\:4n^2-28n-240=0$

$\sf :\implies\:n^2-7n-60=0$

Using Middle Term Split,

$\sf :\implies\:n^2-(12-5)n-60=0$

$\sf :\implies\:n^2-12n+5n-60=0$

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

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

$\sf :\implies\:n-12=0\:and\:n+5=0$

$\sf :\implies\:n=12\:and\:n=-5$

{n cannot be negative} So,

$\bf :\implies\:n=12$

Hence, There are 12 terms in the A.P whose sum is 120.

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