If the (11)th term of an AP is (52) and (d=5), what is (a_2)?
(a_2=a_{11}-9d=52-45=7). When moving backward from a known term, subtract the position gap times (d).
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SubjectsMathematics
TOPIC PRACTICE
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(a_2=a_{11}-9d=52-45=7). When moving backward from a known term, subtract the position gap times (d).
View question detailsThis is an AP with (a=10), (d=3). (a_{12}=10+11\times3=43) cm. In word problems, identify the first term and difference.
View question details(a=200), (d=50), so (a_{15}=200+14\times50=900). The weekly deposit is the term, not the total amount.
View question detailsThis forms an arithmetic progression with first term \(a=18\), common difference \(d=2\), and \(n=20\). Using \(a_n=a+(n-1)d\), \(a_{20}=18+(20-1)\times2=18+38=56\). Therefore, the 20th row has 56 seats. Choosing 58 would result from incorrectly adding the difference one extra time. Exam tip: in the \(n\)th term, the common difference is added \(n-1\) times.
View question detailsThis is an arithmetic progression with first term \(a=25\) cm and common difference \(d=1.5\) cm. The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{30}=25+(30-1)\times1.5=25+43.5=68.5\) cm. Hence, 68.5 cm is correct. The value 70 cm can result from incorrectly adding \(30\times1.5\); the plant grows only 29 times after the first day. Exam tip: use \((n-1)d\), not \(nd\), for the \(n\)th term of an AP.
View question detailsHere (a=500), (d=-20), so (a_{14}=500+13(-20)=240). In decreasing word problems, take (d) as negative.
View question detailsThis is an arithmetic progression with first term \(a=12\) and common difference \(d=16-12=4\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{18}=12+(18-1)\times4=12+68=80\). Hence, the 18th chapter will have 80 pages. Option 84 would result from incorrectly using a larger increase than the actual common difference. Exam tip: Always use \(n-1\), not \(n\), in the nth-term formula.
View question details(a_{11}-a_5=(101)-(47)=54), or (6d=6\times9=54). Use the difference of positions for term differences.
View question detailsIn an AP, the difference between two terms equals the corresponding number of common differences. Thus, \(a_{25}-a_{15}=10d\), so \(120-70=10d\) and \(d=5\). Now, using \(a_{15}=a_1+14d\), we get \(70=a_1+14\times5\), hence \(a_1=0\). Option 5 is the common difference, not the first term. Exam tip: In \(a_n=a_1+(n-1)d\), remember to use \(n-1\).
View question detailsHere, the first term is \(a=x\) and the common difference is \(d=4\). Using \(a_n=a+(n-1)d\), we get \(50=x+(10-1)\times4=x+36\). Hence, \(x=14\). If 16 were chosen, the 10th term would be \(16+36=52\), not 50. Exam tip: In the nth-term formula, use \(n-1\), not \(n\).
View question detailsFrom (-30=27+(n-1)(-3)), (57=3(n-1)), so (n=20). Handle signs carefully with a negative target term.
View question details(d=\frac{64-29}{14-7}=5), hence (a_{21}=64+7\times5=99). Equal position gap (7) gives term gap (35).
View question detailsThe \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{13}=4+(13-1)\times\frac{3}{2}=4+12\times\frac{3}{2}=4+18=22\). Hence, 22 is correct. Getting 21 would mean the common difference has not been multiplied correctly. Exam tip: always use \((n-1)\), not \(n\), in the formula for \(a_n\).
View question detailsGiven \(a_n=50-4n\). For the term whose value is \(-10\), set \(50-4n=-10\). Thus, \(4n=60\), so \(n=15\). Therefore, \(-10\) is the 15th term of the AP. The 14th term is \(50-4(14)=-6\), so it is not correct. Exam tip: To find a term number, equate the given term value to \(a_n\) and solve for \(n\).
View question details(d=\frac{55-31}{8}=3), so (a_6=31-4\times3=19). When moving backward from a known term, subtract (d).
View question detailsThe first term is 8 and the common difference is 6. Using \(a_n=a+(n-1)d\), \(74=8+(n-1)\times6\). Thus, \(66=6(n-1)\), so \(n-1=11\) and \(n=12\). If \(n=13\), the term would be \(8+12\times6=80\), not 74. Exam tip: after finding \(n-1\), remember to add 1 to obtain the term number.
View question detailsSubtracting the relations gives (d=-1), and substitution gives (a_{p+q}=0). Even in symbolic APs, use (a_n=a+(n-1)d).
View question detailsGiven \(a_3=8\) and \(a_8=3\), we have \(a_8-a_3=5d\). Thus, \(3-8=5d\), so \(d=-1\). Now \(a_{11}=a_8+3d=3+3(-1)=0\). Hence, 0 is the correct answer. Option 1 is the value of \(a_{10}\), obtained by counting one term less. Exam tip: When using two known terms of an AP, take the difference of their term numbers to find the number of common differences.
View question detailsIn this AP, (a=105), (d=7), so (a_{50}=105+49\times7=448). In an AP of multiples, choose the first correct multiple.
View question detailsIn (9,18,27,\ldots), (9n<200), so the greatest (n=22) and the term is (198). For multiples, take the largest integer below the limit.
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