# How many terms of the AP: –15, –13, –11,--- are needed to make the sum –55? Explain the reason for double answer. - Mathematics

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How many terms of the AP: –15, –13, –11,--- are needed to make the sum –55? Explain the reason for double answer.

#### Solution

Let n number of terms are needed to make the sum -55

Here, first term (a) = –15

Common difference (d) = –13 + 15 = 2

∵ Sum of n terms of an AP

S_n = n/2[2a + (n - 1)d]

⇒ -55 = n/2[2(-15) + (n - 1)2]  .....[∵ Sn = –55 (Given)]

⇒ –55 = –15M + n(n – 1)

⇒ n2 – 16n + 55 = 0

⇒ n2 – 11n – 5n + 55 = 0  .....[By factorisation method]

⇒ n(n – 11) – 5(n – 11) = 0

⇒ (n – 11)(n – 5) = 0

∴ n = 5, 11

Hence, either 5 or 11 terms are needed to make the sum –55 when n = 5,

AP will be –15, –13, –11, –9, –7

So, resulting sum will be –55 because all terms are negative.

When n = 11,

AP will be –15, –13, –11, –9, –7, –5, –3, –1, 1, 3, 5

So resulting sum will be –55 because the sum of terms 6th to 11th is zero.

Concept: Sum of First n Terms of an A.P.
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#### APPEARS IN

NCERT Mathematics Exemplar Class 10
Chapter 5 Arithematic Progressions
Exercise 5.3 | Q 32 | Page 54
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