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) Consider an arithmetic sequence whose 7th term is 34 and 15th term is 66a) Find the common differenceb) Find the 20th term. |
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Answer» the correct value of common differences in this question is 4 and 20th term is 86 7th term =34, 15th tern =66; a+6d=34; a+14d=66/8d=32; d=32/4=8, a+6d=34 :; a+14d=66/8d=32; d=32/8=4; a+6d=34=a+6(4)=35; a=35-24=10; 25th term; a+( n-1)d=10+( 25-1)4=10+24(4)=10+96=106 20 th term ; a=10, d=4; 10+(19)(4)=10+76=86 c d is 4 and 20th terms is 86 10,14,18,22,26,30,34,38,42,46,50,54,58,62,66 is the arithmetic sequence common difference---4 82 is the 20th term |
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