# If a1, a2, a3, ..., an are in A.P., where ai > 0 for all i, show that 1a1+a2+1a2+a3+...+1an-1+an=n-1a1+an - Mathematics

Sum

If a1, a2, a3, ..., an are in A.P., where ai > 0 for all i, show that 1/(sqrt(a_1) + sqrt(a_2)) + 1/(sqrt(a_2) + sqrt(a_3)) + ... + 1/(sqrt(a_(n - 1)) + sqrt(a_n)) = (n - 1)/(sqrt(a_1) + sqrt(a_n))

#### Solution

Given that a1, a2, a3, ..., an are in A.P.

∴ Common difference d = a2 – a1

= a3 – a2

= a4 – a3

= …

= an – an–1

If a2 – a1 = d then sqrt(a_2^2) - sqrt(a_1^2) = d

⇒ (sqrt(a_1) - sqrt(a_1)) (sqrt(a_2) + sqrt(a_1)) = d  .....[∵ a2 – b2 = (a + b)(a – b)]

⇒ 1/(sqrt(a_1) + sqrt(a_2)) = (sqrt(a_2) - sqrt(a_1))/d

Similarly 1/(sqrt(a_2) + sqrt(a_3)) = (sqrt(a_3) - sqrt(a_2))/d

1/(sqrt(a_3) + sqrt(a_4)) = (sqrt(a_4) - sqrt(a_3))/d

....  ....  ....

1/(sqrt(a_(n - 1)) + sqrt(a_n)) = (sqrt(a_n) - sqrt(a_(n - 1)))/d

Adding the above terms, we get

1/(sqrt(a_1) + sqrt(a_2)) + 1/(sqrt(a_2) + sqrt(a_3)) + 1/(sqrt(a_3) + sqrt(a_4)) + ... + 1/(sqrt(a_(n - 1)) + sqrt(a_n))

= 1/d[sqrt(a_2) - sqrt(a_1) + sqrt(a_3) - sqrt(a_4) - sqrt(a_3) + ... sqrt(a_n) - sqrt(a_(n - 1))]

= 1/d[sqrt(a_n) - sqrt(a_1)]  ....(i)

Now an = a1 + (n – 1)d

⇒ an – a1 = (n – 1)d

⇒ sqrt(a_n^2) - sqrt(a_1^2) = (n – 1)d

⇒ (sqrt(a_n) + sqrt(a_1))(sqrt(a_n) - sqrt(a_1)) = (n – 1)d

⇒ sqrt(a_n) - sqrt(a_1) = ((n - 1)d)/(sqrt(a_n) + sqrt(a_1))

⇒ (sqrt(a_n) - sqrt(a_1))/d = (n - 1)/(sqrt(a_n) + sqrt(a_1))   ....(ii)

From equation (i) and equation (ii) we get

1/(sqrt(a_1) + sqrt(a_2)) + 1/(sqrt(a_2) + sqrt(a_3)) + 1/(sqrt(a_3) + sqrt(a_4)) + ... + 1/(sqrt(a_(n - 1)) + sqrt(a_n)) = (n - 1)/(sqrt(a_n) + sqrt(a_1))

Hence proved.

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Chapter 9: Sequences and Series - Exercise [Page 162]

#### APPEARS IN

NCERT Exemplar Class 11 Mathematics
Chapter 9 Sequences and Series
Exercise | Q 10 | Page 162
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