Find the sum of 1 , 3 , 3 , … … to 12 terms.

We need to find the sum of the series which starts from 1 and the series is up to 12 terms. From the series, it is clearly visible that this is a geometric progression whose common ratio is 3 (as 3 1 = 3 3 ). Let us assume, the first term of the series = a and the common term = r . Hence, the sum of first p terms of geometric progression = a r p - 1 r - 1 if , r > 1 . . . i = a 1 - r p 1 - r if , r < 1 . . . ii Here, r = 3 which is greater than 1 . So, we choose the equation i . a = 1 p = 12 So, the sum of first 12 terms = 1 × 3 12 - 1 3 - 1 = 1 × 728 3 - 1 = 728 × 3 + 1 3 - 1 3 + 1 = 728 × 3 + 1 3 - 1 = 364 × 3 + 1

Find the sum of 1 , 3 , 3 , … … to 12 terms.

We need to find the sum of the series which starts from 1 and the series is up to 12 terms. From the series, it is clearly visible that this is a geometric progression whose common ratio is 3 (as 3 1 = 3 3 ). Let us assume, the first term of the series = a and the common term = r . Hence, the sum of first p terms of geometric progression = a r p - 1 r - 1 if , r > 1 . . . i = a 1 - r p 1 - r if , r < 1 . . . ii Here, r = 3 which is greater than 1 . So, we choose the equation i . a = 1 p = 12 So, the sum of first 12 terms = 1 × 3 12 - 1 3 - 1 = 1 × 728 3 - 1 = 728 × 3 + 1 3 - 1 3 + 1 = 728 × 3 + 1 3 - 1 = 364 × 3 + 1

Find the sum of 1 2 , - 2 , 8 2 , … … to 7 terms.

We need to find the sum of the series which starts from 1 2 and the series is up to 7 terms. From the series, it is clearly visible that this is a geometric progression whose common ratio is - 2 2 Let us assume, the first term of the series = a and the common term = r . Hence, the sum of first p terms of geometric progression = a r p - 1 r - 1 if , r > 1 . . . i = a 1 - r p 1 - r if , r < 1 . . . ii Here, r = - 2 2 which is lesser than 1 . So, we choose the equation i i . a = 1 2 p = 7 So, the sum of first 7 terms = 1 2 × 1 - - 2 2 7 1 - - 2 2 = 1 + 1024 2 2 × 1 + 2 2 = 1 + 1024 2 × 2 2 - 1 7 2 = 4095 - 1022 2 7 2 = 585 - 146 2 2 = 585 2 - 292 2

Find the sum of - 1 3 , 1 2 , - 3 4 , ⋯ ⋯ to 7 terms.

We need to find the sum of the series which starts from - 1 3 and the series is up to 7 terms. From the series, it is clearly visible that this is a geometric progression whose common ratio is - 3 2 (as 1 2 - 1 3 = - 3 4 1 2 ). Let us assume, the first term of the series = a and the common term = r . Hence, the sum of first p terms of geometric progression = a r p - 1 r - 1 if , r > 1 , . . . i = a 1 - r p 1 - r if , r < 1 . . . i i Here, r = - 3 2 which is less than 1 . So, we choose the equation i i . a = - 1 3 p = 7 So, the sum = a 1 - r p 1 - r = - 1 3 × 1 - - 3 2 7 1 - - 3 2 = - 1 + 3 7 2 7 3 × 2 + 3 2 = - 128 + 2187 128 15 2 = - 2 × 2315 15 × 128 = - 463 192

Insert 3 geometric means between 2 1 4 and 4 9 .

We have to insert three geometric means between 2 1 4 = 9 4 and 4 9 . Let the three geometric means between 9 4 and 4 9 be G 1 , G 2 , G 3 . Then 9 4 , G 1 , G 2 , G 3 , 4 9 is a G . P . with the 1 s t term = a = 9 4 and 5 t h term = 4 9 . Let the common ratio be r . The n t h term of any G . P . is T n = a r n - 1 Then, T 5 = 9 4 · r 5 - 1 ⇒ 4 9 = 9 4 · r 4 ⇒ 16 81 = r 4 ⇒ 2 3 4 = r 4 ⇒ r = 2 3 Now, using the above formula, we have: G 1 = a r , G 2 = a r 2 , G 3 = a r 3 ∴ G 1 = 9 4 × 2 3 = 3 2 , G 2 = 9 4 × 2 3 2 = 1 , G 3 = 9 4 × 2 3 3 = 2 3 3 2 , 1 , 2 3 are the three required geometric means.

Insert 5 geometric means between 3 5 9 and 40 1 2 .

We have to insert 5 geometric means between 3 5 9 = 32 9 and 40 1 2 = 81 2 . Let the 5 geometric means between 32 9 and 81 2 be G 1 , G 2 , G 3 , G 4 , G 5 . Then 32 9 , G 1 , G 2 , G 3 , G 4 , G 5 , 81 2 is a G . P . with the 1 s t term = a = 32 9 and 7 t h term = 81 2 . Let the common ratio be r . The n t h term of the G . P . with 1 s t term a and common ratio r can be written as T n = a r n - 1 Then, T 7 = 32 9 · r 7 - 1 ⇒ 81 2 = 32 9 · r 6 ⇒ 729 64 = r 6 ⇒ 3 2 6 = r 6 ⇒ 3 2 = r Using above formula, we have: G 1 = a r , G 2 = a r 2 , G 3 = a r 3 , G 4 = a r 4 , G 5 = a r 5 Then, G 1 = 32 9 × 3 2 = 16 3 , G 2 = 32 9 × 3 2 2 = 8 , G 3 = 32 9 × 3 2 3 = 12 , G 4 = 32 9 × 3 2 4 = 18 , G 5 = 32 9 × 3 2 5 = 27 16 3 , 8 , 12 , 18 , 27 are the five required geometric means.

Insert 6 geometric means between 14 and - 7 64 .

We have to insert 6 geometric means between 14 and - 7 64 . Let these 6 geometric means between 14 and - 7 64 be G 1 , G 2 , G 3 , G 4 , G 5 , G 6 . Then 14 , G 1 , G 2 , G 3 , G 4 , G 5 , G 6 , - 7 64 is a G . P . with the 1 s t term = a = 14 and 8 t h term = - 7 64 . Let the common ratio be r . Now, the n t h term of the G . P . can be written as T n = a r n - 1 Then, T 8 = 14 · r 8 - 1 ⇒ - 7 64 = 14 · r 7 ⇒ - 1 128 = r 7 ⇒ - 1 2 7 = r 7 ⇒ - 1 2 = r Using above formula, we have that: G 1 = a r , G 2 = a r 2 , G 3 = a r 3 , G 4 = a r 4 , G 5 = a r 5 , G 6 = a r 6 Then, G 1 = 14 × - 1 2 = - 7 , G 2 = 14 × - 1 2 2 = 7 2 , G 3 = 14 × - 1 2 3 = - 7 4 , G 4 = 14 × - 1 2 4 = 7 8 , G 5 = 14 × - 1 2 5 = - 7 16 , G 6 = 14 × - 1 2 6 = 7 32 - 7 , 7 2 , - 7 4 , 7 8 , - 7 16 , 7 32 are the six required geometric means.

Sum the following series to infinity, 8 5 , - 1 , 5 8 , … … ∞ .

The given series is 8 5 , - 1 , 5 8 , … … ∞ . Let T n represent the n t h term of the given series. Now, T 2 T 1 = T 3 T 2 = - 5 8 Thus, the given series is a geometric progression with 1 s t term a = 8 5 and common ratio r = - 5 8 . Sum of an infinite G . P . with first term a and the common ratio r is given by S ∞ = a 1 - r ⇒ S ∞ = 8 5 1 + 5 8 ⇒ S ∞ = 64 65 Thus, the sum of the given series is 64 65 .