Question

determine the 2nd term of an ap whose 6th term is 12 and 8th term is 32

Answer 1

Answer:

Let a and d be the first term and the common difference of the given AP respectively. Then,

$\sf\small{ \underline{\underline\red{ \pmb{Given:-}}}}$

$\sf \: a_6 = a+5d=12 \: \: \: \: \: ----------(I)$

And

$\sf a_8 = a+7d=32 \: \: - - - - - - - - - - - (2)$

• Subtracting equation (1) and equation (2)

$\sf2d=20 \\ \\ \sf \implies d= \cancel\frac{20}{2} \\ \\ \sf \implies \red{d=10}$

• Substituting d = 10 in (1),

$\sf \: a+5(10)=12 \\ \\ \sf \implies \: a=12-50 \\ \\ \sf \implies \red{a=-38}$

Therefore, the second term of the AP,

• a = -38
• d = 10

$\sf \: a_2=a+(2 - 1)d \\ \\ \sf \implies \: a_2=a+d \\ \\ \sf \implies \: a_2=(-38)+10 \\ \\\sf \implies \: a_2=-28$

• And nth term of the AP

$\sf \: a+(n-1)d \\ \\ \sf \: =(-38)+(n-1)10 \\ \\ \sf \: =(-38)+5n-10 \\ \\ \sf=5n-48$