Solve the equation: 1 + 6 + 11 + … + x = 148 .

Given that 1 + 6 + 11 + … + x = 148 Let S = 1 + 6 + 11 + . . . + x , which is an A . P . with first term = 1 , common difference = 6 - 1 = 5 and last term = x . Since we know that S n = n 2 a + l , where a , l are the first and last terms of A . P . ⇒ 148 = n 2 1 + x and x = 1 + n - 1 × 5 = 5 n - 4 ⇒ n = x + 4 5 So, 148 × 2 = x + 4 5 1 + x ⇒ 296 × 5 = x 2 + 5 x + 4 ⇒ x 2 + 5 x - 1476 = 0 ⇒ x 2 - 36 x + 41 x - 1476 = 0 ⇒ x x - 36 + 41 x - 36 = 0 ⇒ x + 41 x - 36 = 0 Since, x = - 41 is not possible. Hence, x = 36

Solve the equation: 1 + 6 + 11 + … + x = 148 .

Given that 1 + 6 + 11 + … + x = 148 Let S = 1 + 6 + 11 + . . . + x , which is an A . P . with first term = 1 , common difference = 6 - 1 = 5 and last term = x . Since we know that S n = n 2 a + l , where a , l are the first and last terms of A . P . ⇒ 148 = n 2 1 + x and x = 1 + n - 1 × 5 = 5 n - 4 ⇒ n = x + 4 5 So, 148 × 2 = x + 4 5 1 + x ⇒ 296 × 5 = x 2 + 5 x + 4 ⇒ x 2 + 5 x - 1476 = 0 ⇒ x 2 - 36 x + 41 x - 1476 = 0 ⇒ x x - 36 + 41 x - 36 = 0 ⇒ x + 41 x - 36 = 0 Since, x = - 41 is not possible. Hence, x = 36

If ' S ' is the sum of a finite A.P. whose first term is ' a ' and last term is ' l ' , show that its common difference is equal to l 2 - a 2 2 S - a - l .

Given, Sum of finite terms of an A.P. is S . First term of an A.P.is a . Last term of an A.P. is l . We know that, sum of n terms of an A.P when first and last term of an A.P are given is, S n = n 2 a + l . ⇒ S = n 2 a + l ⇒ 2 S = n a + l ⇒ n = 2 S a + l Now, we now that n th term of an A.P. is a n = a + n - 1 d . ⇒ l = a + 2 S a + l - 1 d ∵ n = 2 S a + l ⇒ l = a 2 + a l a + l + 2 S - a - l a + l d ⇒ l a + l = a 2 + a l + 2 S - a - l d ⇒ a l + l 2 - a 2 - a l = 2 S - a - l d ⇒ l 2 - a 2 = 2 S - a - l d ⇒ d = l 2 - a 2 2 S - a - l Hence, proved.

Find the number of terms of a G.P. whose first term is 3 4 and common ratio is 2 and the last term is 384 .

Given a = 3 4 and r = 2 and t n = 384 The n t h term of an G.P. is given by t n = a r n - 1 Now, 384 = 3 4 ( 2 ) n - 1 512 = ( 2 ) n - 1 ( 2 ) 9 = ( 2 ) n - 1 9 = n – 1 10 = n Therefore , number of terms are 10

If G be the G . M . between a and b , show that 1 G 2 - a 2 + 1 G 2 - b 2 = 1 a b = 1 G 2 .

Given that, G be the G . M . between a and b , then G = a b Now, L.H.S = 1 G 2 - a 2 + 1 G 2 - b 2 = 1 a b - a 2 + 1 a b - b 2 = a b - b 2 + a b - a 2 a b - a 2 a b - b 2 = 2 a b - a 2 - b 2 a ( b - a ) · b ( a - b ) = ( a - b ) 2 a · ( a - b ) · b ( a - b ) = ( a - b ) ( a - b ) a b ( a - b ) 2 = 1 a b = 1 G 2 = R.H.S.

Evaluate n t h term of the series : 2 4 + 4 4 2 + 6 4 3 + 8 4 4 + …

Let S = 2 4 + 4 4 2 + 6 4 3 + 8 4 4 + . . . . . ⇒ S = ∑ r = 1 ∞ 2 r 4 r , where 2 r 4 r is the general term of given series . So, the n t h term of given series is = a n = 2 n 4 n .

E M B I B E

Gamma Question Bank for Engineering Mathematics>Chapter 4 - Sequences and Series>Assignment>Q 16

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Solve the equation: $1 + 6 + 11 + \dots + x = 148$ .

Important Questions on Sequences and Series

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JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 4 - Sequences and Series>Assignment>Q 17

If $' S'$ is the sum of a finite A.P. whose first term is $' a'$ and last term is $\Rightarrow x^{2} - 36 x + 41 x - 1476 = 0$ , show that its common difference is equal to $\frac{l^{2} - a^{2}}{2 S - a - l}$ .