Question

║⊕QUESTION⊕║
A Mathematician who is not also something of a poet will never be a complete mathematician
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CLASS 11
SEQUENCES AND SERIES
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The first and the last terms of an AP are -4 and 16 and the sum of the AP is 7171. Find the number of terms in the AP and the common difference

Answer 1

Step-by-step explanation:

$a = t1 = - 4 \\ tn = 16 \\ sn = 7171$

$tn = a + (n - 1)d \\ 16 = - 4 + dn - d \\ dn - d = 20......(1)$

$sn = \frac{n}{2} (2a + (n - 1)d) \\ 7171 = \frac{n}{2} ( - 8 + 20)......(as \: dn - d = 20) \\ 7171 = 6n \\ n = 1195$

$there \: are \: 1195 \: terms..$

$from \: \: \: \: (1) = > \\ dn - d = 20 \\ d(n -1) = 20 \\ d(1194) = 20 \\ d = 0.01675 \\ \\ offfo \: man \: it \: was \: tremendous \: question \: \\ well \: thanks \: for \: the \: question$

Answer 2

Solution:

Given:

=> First term (a) = -4

=> Last term (l)  = 16

=> Sn = 7171

To Find:

=> Number of terms in the A.P

=> Common difference.

Formula used:

$\sf{\implies S_{n}=\dfrac{n}{2}(a+l)}$

$\sf{\implies T_{n}=a+(n-1)d}$

So, first we will find no. of terms in AP

$\sf{\implies S_{n}=\dfrac{n}{2}(a+l)}$

$\sf{\implies 7171=\dfrac{n}{2}(-4+16)}$

$\sf{\implies 7171=\dfrac{n}{2}(12)}$

$\sf{\implies 7171=6n}$

$\sf{\implies n = \dfrac{7171}{6}}$

$\large{\boxed{\boxed{\green{\sf{\implies n=1195}}}}}$

Hence, number of terms in AP is 1195.

Now, we will find common difference,

$\sf{\implies T_{n}=a+(n-1)d}$

$\sf{\implies 16= -4(1195-1)d}$

$\sf{\implies 20 = 1194d}$

$\sf{\implies d = \dfrac{20}{1194}}$

$\large{\boxed{\boxed{\blue{\sf{\implies d = 0.0167}}}}}$