Find the 18 th and 25 th terms of the sequence defined by a n = n n + 3 , if n ∈ N and n is even : 2 n n 2 + 1 , if n ∈ N and n is odd .

Given, a n = n n + 3 , if n ∈ N and n is even : 2 n n 2 + 1 , if n ∈ N and n is odd . Here, n = 18 , which is even. Therefore, a n = n n + 3 ⇒ a 18 = 18 18 + 3 ⇒ a 18 = 18 × 21 ⇒ a 18 = 378 And n = 25 , which is odd. Therefore, a n = 2 n n 2 + 1 ⇒ a 25 = 25 × 2 25 2 + 1 ⇒ a 25 = 50 625 + 1 ⇒ a 25 = 50 626 ⇒ a 25 = 25 313 Hence, the 18 th term and 25 th of the sequence are 378 and 25 313 respectively.

The 4 th term of a geometric sequence is 2 3 and its seventh term is 16 81 . Find the geometric sequence.

Given, t 4 = 2 3 and t 7 = 16 81 . We know that t n = a r n - 1 where a is the first term and r is common ratio and n is the number of terms. Thus, t 4 = a r 4 - 1 = a r 3 ⇒ a r 3 = 2 3 __________ 1 And t 7 = a r 7 - 1 = a r 6 ⇒ a r 6 = 16 81 _________ 2 Dividing equation 2 by equation 1 , we get, ⇒ a r 6 a r 3 = 16 81 ÷ 2 3 ⇒ r 3 = 16 81 × 3 2 ⇒ r = 8 27 3 ⇒ r = 2 3 Substituting r = 2 3 in equation 1 , we get, a 2 3 3 = 2 3 ⇒ a = 2 3 × 27 8 ⇒ a = 9 4 Thus, the first term is 9 4 and the common ratio is 2 3 . A geometric sequence is given by a , a r , a r 2 , a r 3 , . . . . . . . a r n . Therefore, 9 4 , 9 4 2 3 , 9 4 2 3 2 , 9 4 2 3 3 , 9 4 2 3 4 , . . . . . . . . = 9 4 , 3 2 , 1 , 2 3 , 4 9 , . . . . . . . . Hence, 9 4 , 3 2 , 1 , 2 3 , 4 9 , . . . . . . . . is the required geometric sequence.

ln the division algorithm of polynomials the divisor is x + 2 , quotient is x - 1 and the remainder ts 4 . Find the dividend.

Given, Divisor is b = x + 2 Quotient is q = x + 1 Remainder is r = 4 By Euclid's division lemma, we get a = b q + r If we substitute the values, we get ⇒ a = ( x + 2 ) ( x + 1 ) + 4 ⇒ a = x 2 + x + 2 x + 2 + 4 ⇒ a = x 2 + 3 x + 6 Hence, dividend is x 2 + 3 x + 6

Three numbers are in ratio 2 : 5 : 7 . If the first number, the resulting number on subtraction of 7 form the second number and the third number form an arithmetic sequence, then find the numbers.

Let the three numbers be 2 x , 5 x and 7 x It is given that the numbers 2 x , 5 x − 7 , 7 x are in A.P Therefore, 2 ( 5 x − 7 ) = 2 x + 7 x ⇒ 10 x − 14 = 9 x ⇒ 10 x − 9 x = 14 ⇒ x = 14 So, the numbers are 2 × 14 = 28 , 5 × 14 = 70 , 7 × 14 = 98

The 4 th term of a geometric sequence is 2 3 and its seventh term is 16 81 . Find the geometric sequence.

Given, t 4 = 2 3 and t 7 = 16 81 . We know that t n = a r n - 1 where a is the first term and r is common ratio and n is the number of terms. Thus, t 4 = a r 4 - 1 = a r 3 ⇒ a r 3 = 2 3 __________ 1 And t 7 = a r 7 - 1 = a r 6 ⇒ a r 6 = 16 81 _________ 2 Dividing equation 2 by equation 1 , we get, ⇒ a r 6 a r 3 = 16 81 ÷ 2 3 ⇒ r 3 = 16 81 × 3 2 ⇒ r = 8 27 3 ⇒ r = 2 3 Substituting r = 2 3 in equation 1 , we get, a 2 3 3 = 2 3 ⇒ a = 2 3 × 27 8 ⇒ a = 9 4 Thus, the first term is 9 4 and the common ratio is 2 3 . A geometric sequence is given by a , a r , a r 2 , a r 3 , . . . . . . . a r n . Therefore, 9 4 , 9 4 2 3 , 9 4 2 3 2 , 9 4 2 3 3 , 9 4 2 3 4 , . . . . . . . . = 9 4 , 3 2 , 1 , 2 3 , 4 9 , . . . . . . . . Hence, 9 4 , 3 2 , 1 , 2 3 , 4 9 , . . . . . . . . is the required geometric sequence.

Find the three consecutive terms in an A.P. whose sum is 18 and the sum of their squares is 140 .

We know that the three consecutive terms of an A.P. may be taken as a – d , a , a + d . Given, a – d + a + a + d = 18 ⇒ 3 a = 18 ⇒ a = 18 3 = 6 Also given ( a – d ) 2 + a 2 + ( a + d ) 2 = 140 We know that ( a – b ) 2 = a 2 – 2 a b + b 2 And ( a + b ) 2 = a 2 + 2 a b + b 2 ⇒ a 2 – 2 a d + d 2 + a 2 + a 2 + 2 a d + d 2 = 140 ⇒ 3 a 2 + 2 d 2 = 140 ⇒ 3 ( 6 ) 2 + 2 d 2 = 140 ⇒ 108 + 2 d 2 = 140 ⇒ 2 d 2 = 140 – 108 = 32 ⇒ d 2 = 32 2 = 16 ⇒ d = 4 So, the numbers are ⇒ a – d = 6 – 4 = 2 ⇒ a = 6 ⇒ a + d = 6 + 4 = 10 &there4; The three consecutive numbers are 2 , 6 , 10 .

Find the 18 th and 25 th terms of the sequence defined by a n = n n + 3 , if n ∈ N and n is even : 2 n n 2 + 1 , if n ∈ N and n is odd .

Given, a n = n n + 3 , if n ∈ N and n is even : 2 n n 2 + 1 , if n ∈ N and n is odd . Here, n = 18 , which is even. Therefore, a n = n n + 3 ⇒ a 18 = 18 18 + 3 ⇒ a 18 = 18 × 21 ⇒ a 18 = 378 And n = 25 , which is odd. Therefore, a n = 2 n n 2 + 1 ⇒ a 25 = 25 × 2 25 2 + 1 ⇒ a 25 = 50 625 + 1 ⇒ a 25 = 50 626 ⇒ a 25 = 25 313 Hence, the 18 th term and 25 th of the sequence are 378 and 25 313 respectively.