Let a 1- a 2- a 3- a 4- be an arithmetic progression and g 1- g 2- g 3- g 4- be a geometric progression- If a 1- g 1-1- a 2- g 2-4- a 3- g 3-15 and a 4- g 4-2- thenA- the common ratio of geometric progression is equal to -2-B- the common ratio of geometric progression is equal to -3C- -k-120 ak-960D- - k -120 a k -480

The correct options are B the common ratio of geometric progression is equal to –3 C ∑_k=1^20a_k = 960 Let a and d be first term and common difference of A.P. ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Arithmetic Progression

Let a 1, a 2,...

Question

Let $a_{1}, a_{2}, a_{3}, a_{4}$ …….. be an arithmetic progression and $g_{1}, g_{2}, g_{3}, g_{4}$ …… be a geometric progression. If $a_{1} + g_{1} = 1, a_{2} + g_{2} = 4, a_{3} + g_{3} = 15 a n d a_{4} + g_{4} = 2$ , then

the common ratio of geometric progression is equal to –2.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

the common ratio of geometric progression is equal to –3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$\sum_{k = 1}^{20} a_{k} = 960$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$\sum_{k = 1}^{20} a_{k} = 480$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
B
the common ratio of geometric progression is equal to –3

C
$\sum_{k = 1}^{20} a_{k} = 960$

Let a and d be first term and common difference of A.P. Also, let b and r be the first term and common ratio of G.P.
$∴$ On solving, we get
$a = \frac{1}{2}, b = \frac{1}{2}, d = 5, r = - 3 A l s o, \sum_{k = 1}^{20} a_{k} = 960$

Suggest Corrections

Similar questions

Q. Suppose the sequence

a_{1}, a_{2}, a_{3} . . . . .

is an arithmetic progression of distinct numbers such that the

a_{1}, a_{2}, a_{4}, a_{6} . . . . .

is a geometric progression. The common ration of the geometric progression is:

Q. If

a, a_{1}, a_{2}, . . . . ., a_{2 n}, b

are in arithmetic progression and

a, g_{1}, g_{2}, . . . . g_{2 n}, b

are in geometric progression and

h

is the harmonic mean of

a

and

b

, then

\frac{a_{1} + a_{2 n}}{g_{1} g_{2 n}} + \frac{a_{2} + a_{2 n - 1}}{g_{2} g_{2 n - 1}} + . . . . . + \frac{a_{n} + a_{n + 1}}{g_{n} g_{n + 1}}

is equal to

Q. Let

a_{1}, a_{2}, . . . .

be positive real numbers in geometric progression. For each n, let

A_{n}, G_{n}, H_{a}

be respectively, the arithmetic mean, geometric mean and harmonic mean of

a_{1}, a_{2}, . . . ., a_{n} .

Find an expression for the geometric mean of

G_{1}, G_{2}, . . . ., G_{n}

in terms of

A_{1}, A_{2}, . . . . ., A_{n}, H_{1}, H_{2}, . . . . ., H_{n} .

Q. If

a_{1}, a_{2}, a_{3}, \dots

is an arithmetic progression with common difference 1 and

a_{1} + a_{2} + a_{3} + \dots \dots + a_{98} = 137

, then what is the value of

a_{2} + a_{4} + a_{6} + \dots \dots + a_{98}

? [2 MARKS]

Q. If

a_{1}, a_{2}, a_{3}, \dots, a_{n}

are in geometric progression. Then the given geometric progression is a

Join BYJU'S Learning Program

Let a1, a2, a3, a4 …….. be an arithmetic progression and g1, g2, g3, g4 …… be a geometric progression. If a1+g1=1, a2+g2=4, a3+g3=15 and a4+g4=2, then

The correct options are B the common ratio of geometric progression is equal to –3 C ∑20k=1ak=960 Let a and d be first term and common difference of A.P. Also, let b and r be the first term and common ratio of G.P. ∴ On solving, we get a=12, b=12, d=5, r=−3Also, ∑20k=1ak=960

Let $a_{1}, a_{2}, a_{3}, a_{4}$ …….. be an arithmetic progression and $g_{1}, g_{2}, g_{3}, g_{4}$ …… be a geometric progression. If $a_{1} + g_{1} = 1, a_{2} + g_{2} = 4, a_{3} + g_{3} = 15 a n d a_{4} + g_{4} = 2$ , then