Answered By Leonardo F

First, let's write the formula for the sum of the n terms of an arithmetic sequence (S):

 $S=\frac{\left(t_1+t_n\right)n}{2}$S=(t₁+t_n)n2  

We know that: S=432;n=12. The generic term formula is:

 $t_n=t_1+\left(n-1\right)d$t_n=t₁+(n−1)d 

Hence, substituting in the sum formula:

 $S=\frac{\left(t_1+t_1+\left(n-1\right)d\right)n}{2}$S=(t₁+t₁+(n−1)d)n2  

Since d=6:

 $432=\frac{\left(t_1+t_1+\left(12-1\right)6\right)12}{2}$432=(t₁+t₁+(12−1)6)122  

 $432=\frac{\left(2t_1+11\left(6\right)\right)12}{2}$432=(2t₁+11(6))122  

Solving for t1:

 $t_1=3$t₁=3 

Answered By Leonardo F

Question 2) The generic term formula for a geometric sequence is:

 $t_n=t_1.r^{n-1}$t_n=t₁.rⁿ⁻¹ 

Since r=5 and t1=2, we have:

 $t_7=2\times5^{7-1}=2\times5^6=31250$t₇=2×5⁷⁻¹=2×5⁶=31250