Question

how many terms of the arithmetic series 10 + 8 + 6 +... will make the sum 24?

Answer 1

$10+8+6+---$

$a=10 [first\space term]$

$d=a_2-a_1[common\space difference]$

$=8-10\\=-2\\s_n=\frac{n}{2}(2a+(n-1)d) (sum\space of\space n\space terms)$

We know $s_n=24$

$\therefore 24=\frac{n}{2}(2(10)+(n-1)(-2))\\24=\frac{n}{2}(20-2n+2)\\48=20n-2n^2+2n\\2n^2-22n+48=0\\n^2-11n+24=0\\n^2-8n-3n+24=0\\n(n-8)-3(n-8)=0\\(n-3)(n-8)=0\\(n-3)(n-8)=0\\\implies n-3=0,\space n-8=0\\n=3,\space n=8$

no of terms =3,8