Learn to Create a Robotic Device Using Arduino in the Free Webinar. Register Now
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Free webinar on Robotics (Block Chain) Learn to create a Robotic Device Using Arduino
30th Jan @ 5:00PM for Grade 1 to 10
Arithmeticogeometric Progression Solved Examples As the name suggests, an arithmeticogeometric progression is obtained when the corresponding terms of a geometric and arithmetic progression are multiplied by each other. Mathematically, it can be expressed as xn = bnhn, where xn is the nth term of the arithmeticogeometric progression bn is the nth term of the arithmetic progression hn is the nth term of the geometric progression Illustration: Find the sum of series 1. 2 + 2. 22 + 3. 23 +…+ 100. 2100. Solution: Let us denote the given series by S. Let S = 1.2 + 2.22 + 3.23 +…+ 100.2100 …… (1) ⇒ 2S = 1.22 + 2.23 +…+ 99.2100 + 100.2101 …… (2) ⇒ –S = 1.2 + 1.22 + 1.23 +…+ 1.2100 – 100.2101 ⇒ –S = 1.2 (299–1/(2–1)) – 100.2101 ⇒ S = –2100 + 2 + 100.2101 = 199.2100 + 2. Illustration: Let r = 1/2, consider nrn for increasing value of n and then find its value as n tends to infinity. Solution: We first consider n (1/2)n for increasing values of n. n = 1 we get 1. (1/2)1 = 1/2 = 0.5 n = 2 gives 2 × (1/2)2 = 1/2 = 0.5 n = 3 gives 3 × (1/2)3 = 0.375 n = 10 : 10 (1/2)10 = 0.00976, and so on Thus we observe that as n → ∞ n rn → 0 for |r| < 1. Illustration: Evaluate 1 + 4/5 + 7/52 + 10/53 +…… to infinite terms. Solution: We denote the given sequence by S. Let S = 1 + 4/5 + 7/52 + 10/53 + ……… 1/5 S = 1/5 + 4/52 + 7/53 ……… Subtracting (1–1/5) S = 1 + 3/5 + 3/52 + 3/53 + ……… 4/5 S = 1/ (1–3/5) (As it is infinite G.P.) ⇒ S =25/8 Refer the following video for more on arithmeticogeometric progression Let t1, t2, t3, ……, tm–1, tm, tm+1, be a sequence so that (i) tm+1/tm = tm/tm–1 ……… constant (r) then tp = (t1)rp–1 (ii) tm+1/tm = tm/tm–1 = constant (r) then tp = constant 1 + (constant 2) × rp–1 (iii) If the difference of difference of terms are in G.P. then tp = a + bp + crp–1, where r is the common ratio. Illustration: 7. 14. 33. 88. 251. 738 ………… Solution: We can represent the given sequence in the form of a diagram as shown Remark: First we write down all the numbers in a straight line. Then we find out the difference between every pair of two terms. This is denoted in the second line. Again we find the difference between every pair of consecutive terms in the second line. Now we divide every successive term by the preceding term and see that the ratio in each case is 3. 324/108 = 108/36 = 36/12 = 3 Hence, using the formula listed above ∴ tp = a + bp + c 3p–1 p=1 so t1 = 7 = a + b + c p=2 so t2 = 14 = a + 2b + 3c p=3 so t3 = 33 = a + 3b + 9c Solving, we get a = 3, b = 1, c = 3 ⇒ tp = 3 + p + 3. (3p–1) askIITians offers extensive study material which covers all the important topics of IIT JEE Mathematics. Topics like arithmetic progression, geometric progression and arithmeticogeometric progression have been covered along with numerous solved examples. It is important to have clear fundamentals in order to remain competitive in the JEE. Related resources:
As the name suggests, an arithmeticogeometric progression is obtained when the corresponding terms of a geometric and arithmetic progression are multiplied by each other.
Mathematically, it can be expressed as
xn = bnhn, where
xn is the nth term of the arithmeticogeometric progression
bn is the nth term of the arithmetic progression
hn is the nth term of the geometric progression
Illustration:
Find the sum of series 1. 2 + 2. 22 + 3. 23 +…+ 100. 2100.
Solution: Let us denote the given series by S.
Let S = 1.2 + 2.22 + 3.23 +…+ 100.2100 …… (1)
⇒ 2S = 1.22 + 2.23 +…+ 99.2100 + 100.2101 …… (2)
⇒ –S = 1.2 + 1.22 + 1.23 +…+ 1.2100 – 100.2101
⇒ –S = 1.2 (299–1/(2–1)) – 100.2101
⇒ S = –2100 + 2 + 100.2101 = 199.2100 + 2.
Let r = 1/2, consider nrn for increasing value of n and then find its value as n tends to infinity.
Solution: We first consider n (1/2)n for increasing values of n.
n = 1 we get 1. (1/2)1 = 1/2 = 0.5
n = 2 gives 2 × (1/2)2 = 1/2 = 0.5
n = 3 gives 3 × (1/2)3 = 0.375
n = 10 : 10 (1/2)10 = 0.00976, and so on
Thus we observe that as n → ∞ n rn → 0 for |r| < 1.
Evaluate 1 + 4/5 + 7/52 + 10/53 +…… to infinite terms.
Solution: We denote the given sequence by S.
Let S = 1 + 4/5 + 7/52 + 10/53 + ………
1/5 S = 1/5 + 4/52 + 7/53 ………
Subtracting
(1–1/5) S = 1 + 3/5 + 3/52 + 3/53 + ………
4/5 S = 1/ (1–3/5) (As it is infinite G.P.)
⇒ S =25/8
Refer the following video for more on arithmeticogeometric progression
Let t1, t2, t3, ……, tm–1, tm, tm+1, be a sequence so that
(i) tm+1/tm = tm/tm–1 ……… constant (r)
then tp = (t1)rp–1 (ii) tm+1/tm = tm/tm–1 = constant (r)
then tp = constant 1 + (constant 2) × rp–1
(iii) If the difference of difference of terms are in G.P. then
tp = a + bp + crp–1, where r is the common ratio.
7. 14. 33. 88. 251. 738 …………
Solution: We can represent the given sequence in the form of a diagram as shown
Remark: First we write down all the numbers in a straight line. Then we find out the difference between every pair of two terms. This is denoted in the second line. Again we find the difference between every pair of consecutive terms in the second line. Now we divide every successive term by the preceding term and see that the ratio in each case is 3.
324/108 = 108/36 = 36/12 = 3
Hence, using the formula listed above
∴ tp = a + bp + c 3p–1
p=1 so t1 = 7 = a + b + c
p=2 so t2 = 14 = a + 2b + 3c
p=3 so t3 = 33 = a + 3b + 9c
Solving, we get a = 3, b = 1, c = 3
⇒ tp = 3 + p + 3. (3p–1) askIITians offers extensive study material which covers all the important topics of IIT JEE Mathematics. Topics like arithmetic progression, geometric progression and arithmeticogeometric progression have been covered along with numerous solved examples. It is important to have clear fundamentals in order to remain competitive in the JEE.
Related resources:
Click here for the Complete Syllabus of IIT JEE Mathematics.
Look into the Previous Year Papers with Solutions to get a hint of the kinds of questions asked in the exam.
You can get the knowledge of Useful Books of Mathematics here.
To read more, Buy study materials of Sequences and Series comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
Harmonic Mean Table of Content Formula for...
Examples Based on Relationship between AM, GM and...
Arithmetic Mean between two numbers When three...
Geometric Progression (G.P.) Table of contents...
Relationship between A.M. and G.M. Table of...
Arithmetic Progression (A.P.) Table of contents...
Sum of first n terms of AP A sequence of numbers...
Arithmetic Mean Examples What exactly do you mean...
Weighted Means Let a 1 , a 2 , ……, a...
Basic Principle of AP Progressions and series...
Solved Examples based on GP Solved examples :...
Sum to n terms of Special Series Table of contents...
Solved Examples Based on Harmonic mean...
Sequences Table of contents Meaning of Sequence...
Geometric Progression Geometric Progression as the...
Method of Differences Suppose a 1 , a 2 , a 3 ,...
Download IIT JEE Solved Examples Based on...
Basic Concepts of Sequences and Pattern Table of...
Relation between A.M., G.M. and H.M. Let there are...
Introduction of Sequences and Series Table of...
Harmonic Progression Harmonic progression is an...
Arithmetic-Geometric Progression Suppose a 1 , a 2...
Geometric Mean Geometric Mean between two numbers...
Arithmetic Mean of mth Power The concept of...
Series Table of contents Meaning of Series History...