Question

Q5. N2008/12/03 The first term of an arithmetic progression is 6 and the fifth term is 12. The progression has n terms and the sum of all the terms is 90. Find the value of n.

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Answer 1

Answer:

Sum of $15$ terms is $90$

Step-by-step explanation:

Given the first term $a=6$ and the fifth term $a+4d=12$

that is

$6+4d=12\\\\4d=6\\\\d=\frac{6}{4}=\frac{3}{2}$.

also given        $S_{n}=\frac{n}{2}[2a+(n-1)d]=90$,

we have to find the value of $n$.

so, $S_{n} = \frac{n}{2}[2(6)+(n-1)\frac{3}{2} ]$

          $=\frac{n}{2}[12+\frac{3n}{2}-\frac{3}{2} ]\\\\=6n+\frac{3n^{2} }{4}-\frac{3n}{4} \\\\=90$

rearranging,

$\frac{3n^{2} }{4} -\frac{3n}{4} +\frac{24n}{4} =90\\\\3n^{2} -21n=90*4=360$  (dividing throughout by 3)

$n^{2} -7n=120$

that is $n^{2} -7n-120=0$

now solving it by the quadratic formula,

$n=\frac{-b}{2a}$±$\sqrt{b^{2}-4ac }/2a$   , here $a=1,b=-7,c=-120$

$n=\frac{7}{2}$±$\frac{\sqrt{49+480} }{2}$

  $=\frac{7}{2}$±$\frac{\sqrt{529} }{2}$

  $=\frac{7}{2}$±$\frac{23}{2}$

that is

$n=\frac{7+23}{2}\ or\ n=\frac{7-23}{2} \\n=30/2=15$

hence the answer

thank you