Progressions (Sequences & Series) Test - 4

In order to count the number of terms in the AP, use the short cut:
[(last term – first term)/ common difference] + 1. In this case it would become:
[(130 – 20)/5] +1 = 23. Option (b) is correct.

Since the 8th and the 12th terms of the AP are given as 39 and 59 respectively, the difference
between the two terms would equal 4 times the common difference. Thus we get 4d = 59 – 39 =
20. This gives us d = 5. Also, the 8th term in the AP is represented by a + 7d, we get:
a + 7d = 39 Æ a + 7 × 5 = 39 Æ a = 4. Option (c) is correct.

The number of terms in a series are found by:

Trying Option (a),
We get least term 5 and largest term 30 (since the largest term is 6 times the least term).
The average of the A.P becomes (5 + 30)/2 = 17.5
Thus, 17.5 × n = 105 gives us:
to get a total of 105 we need n = 6 i.e. 6 terms in this A.P. That means the A.P. should look like:
5, _ , _ , _, _, 30.
It can be easily seen that the common difference should be 5. The A.P, 5, 10, 15, 20, 25, 30 fits the
situation.
The same process used for option (b) gives us the A.P. 10, 35, 60. (10 + 35 + 60 = 105) and in the
third option 15, 90 (15 + 90 = 105).
Hence, all the three options are correct.

The difference between the amounts at the end of 4 years and 10 years will be the simple interest
on the initial capital for 6 years.
Hence, 360/6 = 60 =(simple interest.)
Also, the Simple Interest for 4 years when added to the sum gives 1240 as the amount.
Hence, the original sum must be 1000.

Solve this question through trial and error by using values of n from the options:
For 19 terms, the series would be 5 + 8 + 11 + …. + 59 which would give us a sum for the series
as 19 × 32 = 608. The next term (20th term of the series) would be 62. Thus, 608 + 62 = 670
would be the sum to 20 terms. It can thus be concluded that for 20 terms the value of the sum of the
series is not less than 670. Option (a) is correct.

The series would be 5, 20, 80, 320, 1280, 5120, 20480. Thus, there are a total of 7 terms in the
series. Option (c) is correct.

16r⁴ = 81 Æ r⁴ = 81/16 Æ r = 3/2. Thus, 4th term = ar³ = 16 × (3/2)³ = 54. Option (c) is correct

Visualising the squares below 84, we can see that the only way to get the sum of 3 squares as 84
is: 2² + 4² + 8² = 4 + 16 + 64. The largest number is 8. Option (a) is correct.

The series will be 301, 308, …….. 497

We need the sum of the series 20 + 24 + 28 to cross 1000. Trying out the options, we can see that
in 20 years the sum of his savings would be: 20 + 24 + 28 + … + 96. The sum of this series would
be 20 × 58 =1160. If we remove the 20th year we will get the series for savings for 19 years. The
series would be 20 + 24 + 28 + …. 92. Sum of the series would be 1160 – 96 = 1064. If we
remove the 19th year’s savings the savings would be 1064 – 92 which would go below 1000.
Thus, after 19 years his savings would cross 1000. Option (a) is correct.

[Note: To go forward in a G.P. you multiply by the common ratio, to go backward in a G.P. you
divide by the common ratio.]

Since the sum of 5 numbers in AP is 30, their average would be 6. The average of 5 terms in AP is
also equal to the value of the 3rd term (logic of the middle term of an AP). Hence, the third term’s
value would be 6. Option (b) is correct.

Think like this:
The average of the first 4 terms is 7, while the average of the first 8 terms must be 11.
Now visualize this :

Hence, d = 4/2 = 2 [Note: understand this as a property of an A.P.]
Hence, the average of the 6th and 7th terms = 15 and the average of the 8th and 9th term = 19
But this (19) also represents the average of the 16 term A.P.
Hence, required answer = 16 × 19 = 304.