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Sum of n Terms

If the sum of...

Question

If the sum of the first n terms of an AP is $4 n - n^{2}$ what is the first term(that is S $_{1}$ )? What is the sum of first two terms? What is the second term? Similarly, find the $3$ rd, the $10$ th and the $n$ th terms.

t_{1} = 3, S_{2} = 4, t_{2} = 1, t_{3} = 1, t_{10} = - 15, t_{n} = 5 - n

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t_{1} = 3, S_{2} = 4, t_{2} = 1, t_{3} = 1, t_{10} = - 16, t_{n} = 5 - n

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t_{1} = 3, S_{2} = 4, t_{2} = 1, t_{3} = 1, t_{10} = - 16, t_{n} = 5 - 2 n

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None of these

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Solution

The correct option is D None of these
$S_{n} = 4 n - n^{2}$
Putting $n = 1$ , we get
$S_{1} = 4 - 1 = 3$
i.e., $t_{1} = 3$
$S_{2} = 4 (2) - (2)$
$S_{2} = 8 - 4 = 4$
$S_{2} = 4$
$\Rightarrow t_{1} + t_{2} = 4$
$\Rightarrow 3 + t_{2} = 4$
$\Rightarrow t_{2} = 1$
$t_{2} - t_{1} = 1 - 3 = - 2$
$\Rightarrow d = - 2$
Then
$t_{3} = t_{2} + d = 1 - 2 = - 1$
i.e.
$t_{3} = - 1$
$t_{10} = a + 9 d = 3 + 9 (- 2)$
$\Rightarrow t_{10} = - 15$
$t_{n} = a + (n - 1) d = 3 + (n - 1) \times (- 2)$
$t_{n} = 5 - 2 n$
So, option D is correct.

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Similar questions

Q. Let

T_{1}, T_{2}, T_{3}, \dots

be terms of an A.P. If

S_{1} = T_{1} + T_{2} + T_{3} + \dots + T_{n}

and

S_{2} = T_{2} + T_{4} + T_{6} + \dots + T_{n - 1}

, where

n

is odd, then the value of

\frac{S_{1}}{S_{2}}

Q. Let

T_{1}, T_{2}, T_{3}, \dots

be terms of an A.P. If

S_{1} = T_{1} + T_{2} + T_{3} + \dots + T_{n}

and

S_{2} = T_{2} + T_{4} + T_{6} + \dots + T_{n - 1}

, where

n

is odd, then the value of

\frac{S_{1}}{S_{2}}

Q. In an A.P. if

S_{1} = T_{1} + T_{2} + T_{3} + . . . + T_{n}

(n odd),

S_{2} = T_{2} + T_{4} + T_{6} + . . . + T_{n - 1}

, then find the value of

S_{1} / S_{2}

in terms of n

Q. In an A.P., if

S_{1} = T_{1} + T_{2} + T_{3} + . . . . . . + T_{n}

(

n

odd) and

S_{2} = T_{2} + T_{4} + T_{6} + . . . . + T_{n - 1}

, then find the value of

\frac{S_{1}}{S_{2}}

in terms of

n

Q. Assertion :In an

A . P .

of odd number of terms, let

S_{1}

S_{2}

are such that

S_{1} = t_{1} + t_{2} + t_{3} + . . . + t_{n}

and

S_{2} = t_{1} + t_{3} + t_{5} + . . . + t_{n}

then

\frac{s_{1}}{s_{2}} = \frac{n}{n + 1}

Reason: If

1, 2, 3, . . ., n

be the numbers where

n

is odd then

1, 3, 5, 7 . . . n

will be

\frac{n + 1}{2}

odd numbers &

2, 4, 6 . . . . (n - 1)

will be

\frac{n^{2} - 1}{2}

even numbers.

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If the sum of the first n terms of an AP is 4n−n2 what is the first term(that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

The correct option is D None of theseSn=4n−n2Putting n=1, we get S1=4−1=3i.e., t1=3S2=4(2)−(2)S2=8−4=4 S2=4⇒t1+t2=4⇒3+t2=4⇒t2=1t2−t1=1−3=−2⇒d=−2Then t3=t2+d=1−2=−1i.e.t3=−1t10=a+9d=3+9(−2) ⇒t10=−15tn=a+(n−1)d=3+(n−1)×(−2)tn=5−2nSo, option D is correct.

If the sum of the first n terms of an AP is $4 n - n^{2}$ what is the first term(that is S $_{1}$ )? What is the sum of first two terms? What is the second term? Similarly, find the $3$ rd, the $10$ th and the $n$ th terms.