SSC Exam > SSC Questions > If a1, a2 ... are in A.P., then,1/√a1 +√a2 +...
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If a1, a2 ... are in A.P., then,
1/√a1 +√a2 + 1/√a2 + √a3 + …………… 1/√an +√an+1 is equal to
- a)n/√an +√an+1
- b)n-1/√a1 +√an-1
- c)n-1/√a1 +√an
- d)n/√a1 +√an+1
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
If a1, a2 ... are in A.P., then,1/√a1 +√a2 + 1/√a2 + √a3 + …………… 1/√a...
We have, 1/√a1 +√a2 + 1/√a2 + √a3 + …………… 1/√an +√an+
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Now, 1/ √a1 +√a2 = √a2 +√a1 / (√a2+√a1) (√a2+√a1)
(Multiplying numerator and denominator by (√a2 - √a1
= √a2 + √a3 / (a2 – a1)
= √a2 + √a1 /d
(where d is the common difference)
Similarly, 1/ √a2 + √a3 = (√a3 +√a2)/d and so on.
Then the expression 1/√a1 + √a2 +1/√a2+√a3 …….. + 1/√an + √an+1 can be written as
1/d (√a2 - √a1 + √a3 - √a3 + …………. √an+1 - √an)
=n/nd ((√an+1 - (√a1)
= n(√an+1 - √a1)/ an+1 - a1 [(an+1 - an = nd)]
= n/(√a1+ √an+1)
Most Upvoted Answer
If a1, a2 ... are in A.P., then,1/√a1 +√a2 + 1/√a2 + √a3 + …………… 1/√a...
Answer:
To solve this problem, let's first simplify the given expression:
1/√a1 √a2 1/√a2 √a3 …………… 1/√an √an 1
We can rewrite this expression as:
1/√a1 * √a2 * 1/√a2 * √a3 * … * 1/√an * √an * 1
Now, let's observe the pattern in this expression. We can see that each term in the expression is defined by the ratio of two consecutive terms in the arithmetic progression (A.P.).
The first term, 1/√a1, is the ratio of the first term and the second term in the A.P. The second term, √a2, is the second term in the A.P. The third term, 1/√a2, is the ratio of the second term and the third term in the A.P. And so on.
Using the formula for the sum of an arithmetic progression:
The sum of an arithmetic progression with first term a, common difference d, and n terms can be calculated using the formula:
Sn = n/2[2a + (n-1)d]
In this case, the first term is 1/√a1, the common difference is √a2 - 1/√a1, and the number of terms is n. Therefore, we can calculate the sum of the given expression as:
Sn = n/2[2(1/√a1) + (n-1)(√a2 - 1/√a1)]
Simplifying this expression further, we get:
Sn = n/2[2/√a1 + (√a2 - 1/√a1)(n-1)]
Sn = n/√a1 + (√a2 - 1/√a1)(n-1)/2
Comparing the simplified expression with the options:
Now, let's compare the simplified expression with the options given:
a) n/√an √an 1
b) n-1/√a1 √an-1
c) n-1/√a1 √an
d) n/√a1 √an 1
From the simplified expression, we can see that the correct option is (a) n/√an √an 1, as it matches with the simplified expression.
Therefore, the correct answer is option 'A' - n/√an √an 1.
To solve this problem, let's first simplify the given expression:
1/√a1 √a2 1/√a2 √a3 …………… 1/√an √an 1
We can rewrite this expression as:
1/√a1 * √a2 * 1/√a2 * √a3 * … * 1/√an * √an * 1
Now, let's observe the pattern in this expression. We can see that each term in the expression is defined by the ratio of two consecutive terms in the arithmetic progression (A.P.).
The first term, 1/√a1, is the ratio of the first term and the second term in the A.P. The second term, √a2, is the second term in the A.P. The third term, 1/√a2, is the ratio of the second term and the third term in the A.P. And so on.
Using the formula for the sum of an arithmetic progression:
The sum of an arithmetic progression with first term a, common difference d, and n terms can be calculated using the formula:
Sn = n/2[2a + (n-1)d]
In this case, the first term is 1/√a1, the common difference is √a2 - 1/√a1, and the number of terms is n. Therefore, we can calculate the sum of the given expression as:
Sn = n/2[2(1/√a1) + (n-1)(√a2 - 1/√a1)]
Simplifying this expression further, we get:
Sn = n/2[2/√a1 + (√a2 - 1/√a1)(n-1)]
Sn = n/√a1 + (√a2 - 1/√a1)(n-1)/2
Comparing the simplified expression with the options:
Now, let's compare the simplified expression with the options given:
a) n/√an √an 1
b) n-1/√a1 √an-1
c) n-1/√a1 √an
d) n/√a1 √an 1
From the simplified expression, we can see that the correct option is (a) n/√an √an 1, as it matches with the simplified expression.
Therefore, the correct answer is option 'A' - n/√an √an 1.
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If a1, a2 ... are in A.P., then,1/√a1 +√a2 + 1/√a2 + √a3 + …………… 1/√an +√an+1 is equal toa) n/√an +√an+1b) n-1/√a1 +√an-1c) n-1/√a1 +√and) n/√a1 +√an+1Correct answer is option 'A'. Can you explain this answer?
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If a1, a2 ... are in A.P., then,1/√a1 +√a2 + 1/√a2 + √a3 + …………… 1/√an +√an+1 is equal toa) n/√an +√an+1b) n-1/√a1 +√an-1c) n-1/√a1 +√and) n/√a1 +√an+1Correct answer is option 'A'. Can you explain this answer? for SSC 2025 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about If a1, a2 ... are in A.P., then,1/√a1 +√a2 + 1/√a2 + √a3 + …………… 1/√an +√an+1 is equal toa) n/√an +√an+1b) n-1/√a1 +√an-1c) n-1/√a1 +√and) n/√a1 +√an+1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for SSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a1, a2 ... are in A.P., then,1/√a1 +√a2 + 1/√a2 + √a3 + …………… 1/√an +√an+1 is equal toa) n/√an +√an+1b) n-1/√a1 +√an-1c) n-1/√a1 +√and) n/√a1 +√an+1Correct answer is option 'A'. Can you explain this answer?.
If a1, a2 ... are in A.P., then,1/√a1 +√a2 + 1/√a2 + √a3 + …………… 1/√an +√an+1 is equal toa) n/√an +√an+1b) n-1/√a1 +√an-1c) n-1/√a1 +√and) n/√a1 +√an+1Correct answer is option 'A'. Can you explain this answer? for SSC 2025 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about If a1, a2 ... are in A.P., then,1/√a1 +√a2 + 1/√a2 + √a3 + …………… 1/√an +√an+1 is equal toa) n/√an +√an+1b) n-1/√a1 +√an-1c) n-1/√a1 +√and) n/√a1 +√an+1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for SSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a1, a2 ... are in A.P., then,1/√a1 +√a2 + 1/√a2 + √a3 + …………… 1/√an +√an+1 is equal toa) n/√an +√an+1b) n-1/√a1 +√an-1c) n-1/√a1 +√and) n/√a1 +√an+1Correct answer is option 'A'. Can you explain this answer?.
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