Answer:
Find the (r) between adjacent members
a2/a1=+25/5=5
a3/a2=+125/+25=5
a4/a3=+625/+125=5
a5/a4=+3125/+625=5
a6/a5=+15625/+3125=5
The ration (r) between every two adjacent members of the series is constant and equal to 5
General Form: a
n
=a
1
×r
n-1
This geometric series: a
n
=5×5
n-1
The nature of this series
Sum of finite geometric series members
sum of a geometric series
The sum of our particular series
a 1-rn = 5 1-56 = 5 1-15625 = 5 -15624 =5×3906=19530
1-r 1-5 -4 -4
Finding the n
th
element
a2 =a1×r2-1 =a1×r1 =5×51 =25
a3 =a1×r3-1 =a1×r2 =5×52 =125
a4 =a1×r4-1 =a1×r3 =5×53 =625
a5 =a1×r5-1 =a1×r4 =5×54 =3125
a6 =a1×r6-1 =a1×r5 =5×55 =15625
a7 =a1×r7-1 =a1×r6 =5×56 =78125
a8 =a1×r8-1 =a1×r7 =5×57 =390625
a9 =a1×r9-1 =a1×r8 =5×58 =1953125
a10 =a1×r10-1 =a1×r9 =5×59 =9765625
a11 =a1×r11-1 =a1×r10 =5×510 =48828125
a12 =a1×r12-1 =a1×r11 =5×511 =244140625
a13 =a1×r13-1 =a1×r12 =5×512 =1220703125
a14 =a1×r14-1 =a1×r13 =5×513 =6103515625
a15 =a1×r15-1 =a1×r14 =5×514 =30517578125
a16 =a1×r16-1 =a1×r15 =5×515 =152587890625
