Dinesh Pk

MATHS – ARITHMETIC SEQUENCES Prepared By – Dinesh P.K Lecture & Sound – Savitha C.K ( Maths Teacher ) MATHS – ARITHMETIC SEQUENCES Q1. Write down the first three terms of the sequence of natural numbers leaving remainder 1 on division by 5. Check whether 510 is a term of above sequence. Ans : 1, 6, 11…. is the natural number leaving remainder 1 on division by 5. 510 is not a form of this sequence since 510 – 1 = 509 is not divisible by 5. Q2. 98 IS A TERM OF THE ARITHMETIC SEQUENCE HAVING COMMON DIFFERENCE 7. IS 2016 A TERM OF THIS SEQUENCE ? WHY ? ANS : 2016 – 98 =- IS A MULTIPLE OF 7 HENCE 2016 IS A TERM OF THE SEQUENCE. Q3. ARITHMETIC SEQUENCE STARTS AS 5, 9, 13… WHAT IS THE NEXT TERM ? IS 2012 A TERM OF THIS SEQUENCE? WHY ? ANS : GIVEN SEQUENCE IS 5, 9, 13, . . . . . COMMON DIFFERENCE , D = 9 – 5 = 4 THE NEXT TERM = 13 + 4 = 17 TN = A + (N – 1) X D = 5 + (N – 1) X 4 = 5 + 4N – 4 = 1 + 4N IF 1 IS SUBTRACTED FROM ANY TERM OF THIS SEQUENCE, THE RESULTING NUMBER IS A MULTIPLE OF COMMON DIFFERENCE 4. 2012 – 1 = 2011 BUT 2011 IS NOT A MULTIPLE OF 4. HENCE, 2012 IS NOT A TERM OF THIS SEQUENCE. Q4. THE PRODUCT OF FIRST TWO TERMS OF AN ARITHMETIC SEQUENCE WITH COMMON DIFFERENCE 6 IS 135. FIND THE FIRST TERM ? ANS : LET THE TWO TERMS OF AN A.S BE A AND A+6. (AS COMMON DIFFERENCE IS 6). ACCORDING TO QUESTION, SUM = 6 (15 -9) PRODUCT = ( - 135) (15 X – 9) Therefore the first tem is either – 15 or 9. Q5. (i) What is the sum of first 20 natural numbers ? (ii) The algebraic form of an arithmetic sequence is 6n + 5. Find the sum of the first 20 terms of this sequences. Ans : (i) For finding the sum = (ii) For finding x1 = 11 and x20 = 125 For finding the sum = = 10 x 136 = 1360 Q6. Sum of first n terms of an arithmetic sequence is 3n2 + n. Find the first term and common difference of this sequence. Ans : Sum of the first n term is 3n2 + n First term 3 x 1 +1 = 4 Sum of first two term = 3 x 22 + 2 = 14 Second term = 14 – 4 = 10 Common difference = 10 – 4 = 6