6) Consider the Arithmatic Sequence 5, 9, 13.
a) what is the next term ?
b) is 2021 is a term of this Arithmatic Sequence? Why?
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CBSE
Mathematics
Grade 11
Sum of n Terms of an AP
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Consider the arithmetic sequence 9, 15, 21.
(a). Write the algebraic form of this sequence.
(b). Find the twenty-fifth term of this sequence
(c). Find the sum of terms from twenty-fifth to fiftieth of this sequence
(d). Can the sum of some terms of this sequence be 2015? Why?
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Hint: To find each of the answers, we apply the formulae of the sum of n terms in a sequence and the nthterm of the sequence. The algebraic form is expressing the sequence in form of variables. To find if the sum is 2015, we equate the sum of n terms equal to 2015 and solve for n.
Complete step-by-step solution:
Given Arithmetic sequence: 9, 15, 21
The first term a, of this sequence is a = 9 and
The common difference d is,d=15–9=21–15=6.
(a). Algebraic form
We know for a natural number n, the nthterm of a sequence is given by nth term = a+(n−1)d
i.e. Xn = 9+(n−1)6=6n+3, where n = 1, 2, 3…..
Hence algebraic form of the sequence 9, 15, 21 is Xn= 6n + 3
(b). Twenty fifth term of the sequence
We know the nth term of the sequence is given by nth term = a+(n−1)d
25th term = 9+(25–1)6=153
(c). Sum of terms from twenty fifth to fiftieth.
We know the sum of first n terms of a sequence is given by, Sn=12n[X1+Xn], where X1 is the first term of the sequence and Xn is the nth term of the sequence.
(Sum of terms from twenty-fifth to fiftieth, Sum of first fifty terms subtracted by the sum of first twenty-four terms.
The 24thterm of the sequence is given by, 9+(24–1)6=147
The 50thterm of the sequence is given by, 9+(50–1)6=303
Sum of the first 24 terms, S24=242[X1+X24]=12[9+147]=1872
Sum of the first 50 terms, S50=502[X1+X50]=25[9+303]=7800
Sum of the terms from twenty fifth to fiftieth
= S50−S24
= 7800–1872
= 5928.
(d). Sum of some terms of this sequence be 2015?
Let the sum of the first n terms of the sequence be 2015, Sn=12n[X1+Xn]
⇒2015=12n[9+6n+3]
⇒2015=3n2+6n
⇒3n2+6n−2015 ……… (i)
Now, using the formula for finding the roots of a quadratic equation ax2+bx+c=0 is given by, −b±b2−4ac−−−−−−−√2a, provided b2−4ac⩾0
Comparing equation (i), with this, we get a = 3, b = 6 and c = -2015.
n=−6±62−4×3×(−2015)−−−−−−−−−−−−−−−−−√2×3
n=−6±24216−−−−−√6